CBSE Class 12 Physics Atoms And Nuclie Boards Questions Worksheet

Question 801. Define the distance of closest approach. 
Answer:
The distance of closest approach is defined as the minimum distance to which a high-speed alpha particle can travel directly toward the center of a target nucleus before its entire kinetic energy is converted into electrostatic potential energy, causing it to come to a temporary stop and retrace its path back.
The formula for the distance of closest approach is given by: \[ r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{E_K} \] For a typical \( 5.5\text{ MeV} \) alpha particle incident on a gold nucleus, this value is approximately: \[ r_0 \approx 2.5 \times 10^{-14}\text{ m} \]
In simple words: It is the closest distance an alpha particle can get to a nucleus before the positive charges repel each other so strongly that the particle has to stop and turn back.

Exam Tip: Always state that at the distance of closest approach, the initial kinetic energy of the alpha particle is completely converted into electrical potential energy of the system.

Question 802. The K. E. of \alpha -particle incident on gold foil is doubled. How does the distance of closest approach change?
Answer:
The distance of closest approach \( r_0 \) is related to the kinetic energy \( E_K \) of the incident alpha particle by the formula: \[ r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{E_K} \]
\( \implies r_0 \propto \frac{1}{E_K} \)
When the kinetic energy of the alpha particle is doubled (\( E_K' = 2E_K \)), the new distance of closest approach becomes: \[ r_0' = \frac{r_0}{2} \]
Hence, the distance of closest approach is halved when the kinetic energy is doubled.
In simple words: If you double the starting energy of the particle, it has more strength to push against the repulsion, allowing it to get twice as close to the nucleus.

Exam Tip: Clearly show the inverse proportionality relation between \( r_0 \) and \( E_K \) in your steps to secure full marks.

Question 803. In the Rutherford’s scattering experiment the distance of closest approach for an \alpha particle is \( d_0 \). If \alpha -particle is replaced by a proton, how much kinetic energy in comparison to \alpha particle will it require to have the same distance of closest approach \( d_0 \)? 
Answer:
Let \( q_{\alpha} = 2e \) be the charge of the alpha particle and \( q_p = e \) be the charge of the proton. The formula for the distance of closest approach is: \[ d_0 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{E_K} \]
For the alpha particle: \[ E_{K\alpha} = \frac{1}{4\pi\varepsilon_0} \frac{Ze \cdot (2e)}{d_0} \]
For the proton: \[ E_{Kp} = \frac{1}{4\pi\varepsilon_0} \frac{Ze \cdot e}{d_0} \]
Dividing the equation for the proton's kinetic energy by the equation for the alpha particle's kinetic energy: \[ \frac{E_{Kp}}{E_{K\alpha}} = \frac{Ze \cdot e}{Ze \cdot (2e)} = \frac{1}{2} \]
\( \implies E_{Kp} = \frac{1}{2} E_{K\alpha} \)
Thus, the proton will require only half the kinetic energy of the alpha particle to reach the same distance of closest approach.
In simple words: Since a proton has only half the positive charge of an alpha particle, it experiences half as much repelling force. Therefore, it needs only half the starting energy to reach the same distance.

Exam Tip: Be careful to distinguish between the charge of an alpha particle (\( 2e \)) and a proton (\( e \)) when setting up the potential energy terms.

Question 804. Determine the distance of closest approach when an alpha particle of kinetic energy 4.5 MeV strikes a nucleus of Z = 80, stops and reverses its direction. 
Answer:
The kinetic energy of the alpha particle is: \[ E_K = 4.5\text{ MeV} = 4.5 \times 10^6 \times 1.6 \times 10^{-19}\text{ J} = 7.2 \times 10^{-13}\text{ J} \]
The atomic number of the target nucleus is \( Z = 80 \). The expression for the distance of closest approach is: \[ r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{E_K} \]
Substituting the known values (\( \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2 \) and \( e = 1.6 \times 10^{-19}\text{ C} \)): \[ r_0 = 9 \times 10^9 \times \frac{2 \times 80 \times (1.6 \times 10^{-19})^2}{7.2 \times 10^{-13}} \] \[ r_0 = \frac{9 \times 10^9 \times 160 \times 2.56 \times 10^{-38}}{7.2 \times 10^{-13}} \] \[ r_0 = 5.12 \times 10^{-14}\text{ m} \]
In simple words: Using the conservation of energy formula, we find that the alpha particle can get as close as \( 5.12 \times 10^{-14}\text{ meters} \) to the heavy nucleus before turning around.

Exam Tip: Make sure to convert the energy from MeV to Joules before substituting it into the standard SI formula to prevent calculation errors.

Question 805. (i) What is Impact parameter ? 
(ii) What is the significance of impact parameter ?
Answer:
(i) **Impact Parameter (\( b \)):**
The impact parameter is defined as the perpendicular distance of the initial velocity vector of the incident alpha particle from the center line of the target nucleus when the particle is far away.
(ii) **Significance:**
By measuring the distribution of scattering angles for different impact parameters, the experiment provides a reliable estimate of the physical size of the nucleus.
In simple words: (i) It is the straight-line distance by which an incoming alpha particle would miss the center of the nucleus if it were not deflected. (ii) It helps scientists estimate the actual size of the nucleus.

Exam Tip: A simple, labeled diagram showing the perpendicular distance \( b \) relative to the nucleus center line helps secure complete marks.

Question 806. The trajectories, traced by different \alpha -particles, in Geiger-Marsden experiment were observed as shown in figure. (a) What names are given to the symbols 'b' and '\theta' shown here ? 
(b) What can we say about values of b for (i) \theta =0^0 (ii) \theta = \pi radians ?
Answer:
(a) The symbol \( b \) is the **impact parameter**, and \( \theta \) is the **scattering angle** of the alpha particle.

(b) The relationship between the impact parameter \( b \) and the scattering angle \( \theta \) is given by: \[ b = \frac{Ze^2 \cot(\theta/2)}{4\pi\varepsilon_0 \left(\frac{1}{2}mv^2\right)} \]
(i) When \( \theta = 0^\circ \), \( \cot(0^\circ) \to \infty \), meaning the value of \( b \) is **maximum**. This corresponds to particles that are very far from the nucleus, which pass through undeflected and help estimate the overall atomic size.

(ii) When \( \theta = \pi\text{ radians} \) (\( 180^\circ \)), \( \cot(90^\circ) = 0 \), meaning the value of \( b \) is **zero** (minimum). This corresponds to a head-on collision, where the alpha particle is repelled straight back, helping to determine the nuclear boundary size.
In simple words: (a) \( b \) is the sideways distance from the center, and \( \theta \) is the angle the particle is deflected. (b) A sideways distance of zero makes the particle bounce straight back at 180 degrees, while a very large sideways distance means it flies straight past with zero deflection.

Exam Tip: Remember that a head-on collision (\( b = 0 \)) results in backward scattering (\( \theta = 180^\circ \)), which is the direct evidence for the nuclear model.

Question 807. State Bohr’s quantization condition for defining stationary orbits. 
Answer:
Bohr's quantization condition states that an electron can revolve around the nucleus only in specific non-radiating, stable circular paths known as stationary orbits. In these allowed orbits, the orbital angular momentum \( L \) of the electron must be an integral multiple of \( \frac{h}{2\pi} \): \[ mvr = n\frac{h}{2\pi} \] where \( m \) is the mass of the electron, \( v \) is its velocity, \( r \) is the radius of the orbit, \( h \) is Planck's constant, and \( n \) is the principal quantum number (\( n = 1, 2, 3, \dots \)).
While moving in these quantized orbits, electrons do not lose energy by electromagnetic radiation.
In simple words: Electrons can only orbit the nucleus in specific fixed paths where their angular momentum is a whole-number multiple of a basic value. In these paths, they do not lose energy.

Exam Tip: Be sure to write the formula \( mvr = n\frac{h}{2\pi} \) and define what each variable (\( m, v, r, n, h \)) stands for.

Question 908. State Bohr postulate of hydrogen atom that gives the relationship for the frequency of emitted photon in a transition. 
OR
State Bohr’s postulate of hydrogen atom which successfully explains emission lines in the spectrum of hydrogen atom. 
Answer:
Bohr's transition postulate states that an electron can transition from a higher-energy stationary orbit (energy \( E_2 \)) to a lower-energy stationary orbit (energy \( E_1 \)). During this downward transition, the excess energy is released in the form of a single photon of electromagnetic radiation. The frequency \( \nu \) of the emitted photon is directly determined by the energy difference between these two states: \[ h\nu = E_2 - E_1 \] where \( h \) is Planck's constant. This expression is referred to as Bohr's frequency condition.
In simple words: When an electron jumps down from a high-energy orbit to a lower one, it releases a packet of light (a photon). The energy of this light matches the difference in energy between the two orbits.

Exam Tip: State the frequency condition formula clearly and explain that it forms the physical foundation for the discrete emission lines in the hydrogen spectrum.

Question 809. The ground state energy of hydrogen atom is \( -13.6\text{ eV} \). What are the kinetic and potential energies of electron in this state ? 
Answer:
Let \( E \) be the total energy of the electron in the ground state: \[ E = -13.6\text{ eV} \]
The kinetic energy \( E_K \) is equal to the negative of the total energy: \[ E_K = -E = -(-13.6\text{ eV}) = +13.6\text{ eV} \]
The potential energy \( U \) is equal to twice the total energy: \[ U = 2E = 2 \times (-13.6\text{ eV}) = -27.2\text{ eV} \]
Thus, in this state, the kinetic energy is \( +13.6\text{ eV} \) and the potential energy is \( -27.2\text{ eV} \).
In simple words: In the lowest energy level, the electron's motion gives it \( +13.6\text{ eV} \) of kinetic energy, while the attractive force of the nucleus gives it \( -27.2\text{ eV} \) of potential energy.

Exam Tip: Memorize the simple ratios: \( \text{KE} = -E \) and \( \text{PE} = 2E \). This makes solving these energy-relation questions very quick and reliable.

Question 810. The total energy of an electron in the first excited state of hydrogen atom is \( -3.4\text{ eV} \). What is the kinetic and potential energy of the electron in this state ? 
Answer:
The total energy of the electron in the first excited state is: \[ E = -3.4\text{ eV} \]
The kinetic energy \( E_K \) is: \[ E_K = -E = -(-3.4\text{ eV}) = +3.4\text{ eV} \]
The potential energy \( U \) is: \[ U = 2E = 2 \times (-3.4\text{ eV}) = -6.8\text{ eV} \]
Thus, the kinetic energy is \( +3.4\text{ eV} \) and the potential energy is \( -6.8\text{ eV} \).
In simple words: In the second energy level, the electron has \( +3.4\text{ eV} \) of motion-based energy and \( -6.8\text{ eV} \) of stored electric energy.

Exam Tip: Be sure to write down the positive sign for kinetic energy and the negative sign for potential energy, as signs are critical for grading.

Question 811. Given the value of the ground state energy of hydrogen atom as \( -13.6\text{ eV} \). Find out its kinetic and potential energy in the ground and second excited states. 
Answer:
The total energy in any orbit \( n \) is: \[ E_n = -\frac{13.6}{n^2}\text{ eV} \]
**For the Ground State (\( n = 1 \)):**
Total energy: \[ E_1 = -13.6\text{ eV} \]
Kinetic energy: \[ E_{K1} = -E_1 = +13.6\text{ eV} \]
Potential energy: \[ U_1 = 2E_1 = 2 \times (-13.6\text{ eV}) = -27.2\text{ eV} \]

**For the Second Excited State (\( n = 3 \)):**
Total energy: \[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51\text{ eV} \]
Kinetic energy: \[ E_{K3} = -E_3 = +1.51\text{ eV} \]
Potential energy: \[ U_3 = 2E_3 = 2 \times (-1.51\text{ eV}) = -3.02\text{ eV} \]
In simple words: In the ground state (\( n=1 \)), the kinetic and potential energies are \( +13.6\text{ eV} \) and \( -27.2\text{ eV} \). In the second excited state (\( n=3 \)), they are \( +1.51\text{ eV} \) and \( -3.02\text{ eV} \).

Exam Tip: Remember that the "second excited state" corresponds to \( n = 3 \), not \( n = 2 \). This is a very common trap where students lose marks.

Question 812. The value of ground state energy of hydrogen atom is \( -13.6\text{ eV} \).
(i) what does the negative sign signify ?
(ii) How much energy is required to take an electron in this atom from the ground state to the first excited state ?
Answer:
(i) The negative sign indicates that the electron is bound to the nucleus by an attractive electrostatic force. It means that energy must be supplied to the system from outside to remove the electron to an infinite distance where it becomes free.

(ii) The energy of the electron in any state \( n \) is: \[ E_n = -\frac{13.6}{n^2}\text{ eV} \]
For the ground state (\( n = 1 \)): \[ E_1 = -13.6\text{ eV} \]
For the first excited state (\( n = 2 \)): \[ E_2 = -\frac{13.6}{2^2} = -3.4\text{ eV} \]
The energy \( \Delta E \) required for this transition is the difference between these two levels: \[ \Delta E = E_2 - E_1 = -3.4\text{ eV} - (-13.6\text{ eV}) = 10.2\text{ eV} \]
Therefore, \( 10.2\text{ eV} \) of energy is required.
In simple words: (i) The negative sign means the electron is trapped inside the atom by the positive charge of the nucleus. (ii) To push the electron from the lowest level to the next level up, you must supply exactly \( 10.2\text{ eV} \) of energy.

Exam Tip: When defining the significance of the negative sign, use keywords like "bound state" and "attractive electrostatic force" to secure full marks.

Question 813. In the ground state of hydrogen atom, its Bohr radius is given as \( 5.3 \times 10^{-11}\text{ m} \). The atom is excited such that the radius becomes \( 21.2 \times 10^{-11}\text{ m} \). Find - 
(i) the value of principal quantum number and
(ii) the total energy of the atom in this excited state.
Answer:
Let \( r_0 = 5.3 \times 10^{-11}\text{ m} \) be the Bohr radius of the ground state and \( r_n = 21.2 \times 10^{-11}\text{ m} \) be the radius of the excited state.
The radius of the \( n \)-th Bohr orbit is given by: \[ r_n = n^2 r_0 \]
(i) To find the principal quantum number \( n \): \[ n^2 = \frac{r_n}{r_0} = \frac{21.2 \times 10^{-11}\text{ m}}{5.3 \times 10^{-11}\text{ m}} = 4 \]
\( \implies n = 2 \)
Thus, the principal quantum number is \( 2 \).

(ii) The total energy of the atom in the \( n = 2 \) state is: \[ E_2 = -\frac{13.6}{2^2} = -3.4\text{ eV} \]
The total energy in this state is \( -3.4\text{ eV} \).
In simple words: (i) Since the orbit's radius is four times larger than the starting radius, the electron must be in the second level (\( n = 2 \)). (ii) The total energy in this second level is \( -3.4\text{ eV} \).

Exam Tip: Be sure to write the formula \( r_n = n^2 r_0 \) before doing the division to make your derivation steps clear.

Question 814. Calculate the de-Broglie wavelength of the electron orbiting in the \( n = 2 \) state of hydrogen atom.
Answer:
The total energy of the electron in the \( n = 2 \) state is: \[ E_K = -\left(-\frac{13.6}{n^2}\right) = \frac{13.6}{4}\text{ eV} = 3.4\text{ eV} \] Converting this energy into Joules: \[ E_K = 3.4 \times 1.6 \times 10^{-19}\text{ J} = 5.44 \times 10^{-19}\text{ J} \]
The de-Broglie wavelength \( \lambda \) of the electron is: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m E_K}} \] Substituting the mass of the electron \( m = 9.1 \times 10^{-31}\text{ kg} \) and Planck's constant \( h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} \): \[ \lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5.44 \times 10^{-19}}} \] \[ \lambda \approx \frac{6.6 \times 10^{-34}}{9.94 \times 10^{-25}} \approx 0.66 \times 10^{-9}\text{ m} = 0.66\text{ nm} \]
In simple words: By finding the energy of the electron in its second orbit, we can calculate its wave-like wavelength, which turns out to be about \( 0.66\text{ nm} \).

Exam Tip: You can also solve this quickly using Bohr's angular momentum quantization \( 2\pi r = n\lambda \). For \( n=2 \), \( \lambda = \pi r \). If you have the radius, this provides a quick way to double-check your answer.

Question 815. What is the longest wavelength of photon that can ionize a hydrogen atom in its ground state ? Specify the type of radiation.
Answer:
To ionize a hydrogen atom from its ground state, the energy of the incident photon must be at least equal to the ionization energy of the ground state: \[ \Delta E = 13.6\text{ eV} = 13.6 \times 1.6 \times 10^{-19}\text{ J} = 2.176 \times 10^{-18}\text{ J} \]
The relationship between energy and wavelength \( \lambda \) is: \[ \Delta E = \frac{hc}{\lambda} \]
\( \implies \lambda = \frac{hc}{\Delta E} \)
Substituting the values (\( h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} \), \( c = 3 \times 10^8\text{ m/s} \)): \[ \lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.176 \times 10^{-18}} \approx 9.1 \times 10^{-8}\text{ m} = 91\text{ nm} \]
This wavelength of \( 91\text{ nm} \) lies in the **ultraviolet** region of the electromagnetic spectrum.
In simple words: The minimum energy needed to knock the electron out of the atom is \( 13.6\text{ eV} \). The longest light wave that carries this much energy has a wavelength of \( 91\text{ nm} \), which is ultraviolet light.

Exam Tip: Be sure to specify the spectral region (Ultraviolet) as requested in the second part of the question.

Question 816. Write the expression for Bohr’s radius in hydrogen atom. 
Answer:
The expression for the radius of the innermost orbit (\( n = 1 \)) of a hydrogen atom, known as the Bohr radius \( r_0 \), is: \[ r_0 = \frac{\varepsilon_0 h^2}{\pi m e^2} \] where \( \varepsilon_0 \) is the permittivity of free space, \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( e \) is the elementary charge of the electron. Substituting the standard values yields: \[ r_0 \approx 0.53\text{ \AA} = 5.3 \times 10^{-11}\text{ m} \]
In simple words: The formula shows how the radius of the closest orbit depends on physical constants, giving a value of about half an angstrom.

Exam Tip: Write down what each symbol (\( \varepsilon_0, h, m, e \)) represents to ensure a complete answer.

Question 817. In hydrogen atom, if the electron is replaced by a particle which is 200 times heavier but have the same charge, How would its radius change ?
Answer:
The radius of the \( n \)-th Bohr orbit is given by: \[ r = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2} \] This formula shows that the radius of the orbit is inversely proportional to the mass \( m \) of the orbiting particle:
\( \implies r \propto \frac{1}{m} \)
If the electron is replaced by a particle that has the same charge but is 200 times heavier (\( m' = 200m \)), the new radius \( r' \) of the orbit will be: \[ r' = \frac{r}{200} \]
Thus, the radius of the orbit will decrease to **\( \frac{1}{200} \)** of its original value.
In simple words: Since a heavier particle is harder to keep in a wide circle with the same force, the electrostatic pull drags it much closer, making the orbit 200 times smaller.

Exam Tip: Highlight the inverse relationship \( r \propto \frac{1}{m} \) clearly to show the physical reasoning behind your answer.

Question 818. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom ? 
Answer:
The radius \( r_n \) of the \( n \)-th orbit of a hydrogen atom is proportional to the square of the principal quantum number: \[ r_n \propto n^2 \]
For the ground state, the quantum number is \( n_1 = 1 \).
For the first excited state, the quantum number is \( n_2 = 2 \).
Taking the ratio of their radii: \[ \frac{r_2}{r_1} = \left(\frac{n_2}{n_1}\right)^2 = \left(\frac{2}{1}\right)^2 = \frac{4}{1} \]
Thus, the ratio of the radii of the first excited state to the ground state is **4 : 1**.
In simple words: Since orbit radius grows with the square of the level number, the second orbit is \( 2^2 = 4 \) times larger than the first orbit.

Exam Tip: Be careful not to confuse the first excited state (\( n=2 \)) with the first orbit (\( n=1 \)).

Question 819. The radius of innermost electron orbit of a hydrogen atom is \( 5.3 \times 10^{-11}\text{ m} \). What is the radius of orbit in the second excited state ? 
Answer:
The radius of the innermost orbit (\( n = 1 \)) is: \[ r_1 = 5.3 \times 10^{-11}\text{ m} \]
The "second excited state" corresponds to the third orbit, where \( n = 3 \).
The radius of the \( n \)-th orbit is given by: \[ r_n = n^2 r_1 \]
Substituting \( n = 3 \) into the formula: \[ r_3 = 3^2 \times r_1 = 9 \times 5.3 \times 10^{-11}\text{ m} \] \[ r_3 = 47.7 \times 10^{-11}\text{ m} = 4.77 \times 10^{-10}\text{ m} \]
Thus, the radius of the orbit in the second excited state is **\( 4.77 \times 10^{-10}\text{ m} \)**.
In simple words: The second excited state is the third level. Since the radius scales as \( n^2 \), the third orbit is 9 times larger than the first, giving \( 47.7 \times 10^{-11}\text{ meters} \).

Exam Tip: Always double check that you are using \( n = 3 \) for the "second excited state".

Question 820. Find out the wavelength of the electron orbiting in the ground state of hydrogen atom.
Answer:
For the ground state (\( n = 1 \)), the Bohr radius is: \[ r_1 = 0.53\text{ \AA} = 0.53 \times 10^{-10}\text{ m} \]
According to de-Broglie's relation for an electron in a circular Bohr orbit: \[ 2\pi r = n\lambda \] For the ground state (\( n = 1 \)): \[ 2\pi r_1 = \lambda \]
Substituting the value of \( r_1 \): \[ \lambda = 2 \times 3.14 \times 0.53 \times 10^{-10}\text{ m} \] \[ \lambda \approx 3.32 \times 10^{-10}\text{ m} = 3.32\text{ \AA} \]
Thus, the de-Broglie wavelength of the ground-state electron is **\( 3.32\text{ \AA} \)**.
In simple words: The orbit circumference must fit exactly one full wavelength of the electron. Calculating the circumference of the first orbit gives a wavelength of \( 3.32\text{ \AA} \).

Exam Tip: Using the circular wave relation \( 2\pi r = n\lambda \) is the most direct way to solve this transition-wavelength problem.

Question 821. Use Bohr model of hydrogen atom to calculate the speed of the electron in the first excited state.
Answer:
The speed of an electron in the \( n \)-th orbit of a hydrogen atom is given by: \[ v_n = \frac{c}{137} \frac{1}{n} \] where \( c \) is the speed of light in vacuum (\( c \approx 3 \times 10^8\text{ m/s} \)).
For the first excited state, the quantum number is \( n = 2 \). Substituting this value into the speed formula: \[ v_2 = \frac{3 \times 10^8}{137 \times 2}\text{ m/s} \] \[ v_2 = \frac{3 \times 10^8}{274}\text{ m/s} \approx 1.09 \times 10^6\text{ m/s} \]
Thus, the speed of the electron in the first excited state is **\( 1.09 \times 10^6\text{ m/s} \)**.
In simple words: In the second orbit, the electron travels at a high speed of about \( 1.09 \times 10^6\text{ meters per second} \), which is roughly 1/274th the speed of light.

Exam Tip: Remember that speed decreases as \( \frac{1}{n} \), so the electron moves slower in excited states than in the ground state.

Question 822. Use Rydberg formula to determine the wavelength of \( H_\alpha \) line. (Given : Rydberg’s constant \( R = 1.03 \times 10^7\text{ m}^{-1} \)) 
Answer:
The \( H_\alpha \) spectral line of the Balmer series is produced when an electron transitions from the orbit \( n_2 = 3 \) to \( n_1 = 2 \). The Rydberg formula is: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Substituting the given values: \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \] \[ \frac{1}{\lambda} = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36} \]
\( \implies \lambda = \frac{36}{5R} \)
Substituting \( R = 1.03 \times 10^7\text{ m}^{-1} \): \[ \lambda = \frac{36}{5 \times 1.03 \times 10^7} = \frac{36}{5.15 \times 10^7} \approx 6.99 \times 10^{-7}\text{ m} = 6990\text{ \AA} \]
Thus, the wavelength of the \( H_\alpha \) line is approximately **\( 6990\text{ \AA} \)**.
In simple words: When the electron drops from level 3 to level 2, it emits red light with a wavelength of \( 6990\text{ \AA} \).

Exam Tip: Be sure to use the value of Rydberg's constant provided in the question, as it may vary slightly from the standard value of \( 1.097 \times 10^7\text{ m}^{-1} \).

Question 823. When \( H_\alpha \) line in the emission spectrum of hydrogen atom obtained ? Calculate the frequency of photon emitted during this transition. 
Answer:
The \( H_\alpha \) line (the first line of the Balmer series) is obtained when an electron transitions from the \( n = 3 \) orbit to the \( n = 2 \) orbit.
The wavelength of this transition is given by the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5R}{36} \] Using the standard value of Rydberg's constant \( R = 1.09 \times 10^7\text{ m}^{-1} \): \[ \frac{1}{\lambda} = \frac{5 \times 1.09 \times 10^7}{36} \]
The frequency \( \nu \) of the emitted photon is related to wavelength by the equation: \[ \nu = \frac{c}{\lambda} = c \times \left( \frac{1}{\lambda} \right) \] Substituting the speed of light \( c = 3 \times 10^8\text{ m/s} \): \[ \nu = 3 \times 10^8 \times \frac{5 \times 1.09 \times 10^7}{36} \] \[ \nu = \frac{15 \times 1.09 \times 10^{15}}{36} = \frac{1.635 \times 10^{16}}{36} \approx 4.54 \times 10^{14}\text{ Hz} \]
In simple words: This line is created when an electron falls from the 3rd orbit to the 2nd orbit. The light wave emitted has a frequency of about \( 4.54 \times 10^{14}\text{ Hz} \).

Exam Tip: State both the transition levels (\( 3 \to 2 \)) and the calculated frequency to answer both parts of the prompt.

Question 824. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. (Rydberg constant, \( R = 10^7\text{ m}^{-1} \)) 
Answer:
For the Balmer series, the lower level is \( n_1 = 2 \). The shortest wavelength (series limit) corresponds to the transition from the highest possible energy state, which is \( n_2 = \infty \). Applying the Rydberg formula: \[ \frac{1}{\lambda_{\min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
\( \implies \lambda_{\min} = \frac{4}{R} \)
Substituting \( R = 10^7\text{ m}^{-1} \): \[ \lambda_{\min} = \frac{4}{10^7}\text{ m} = 4 \times 10^{-7}\text{ m} = 4000\text{ \AA} \]
Thus, the shortest wavelength of the Balmer series is **\( 4000\text{ \AA} \)**.
In simple words: The shortest wavelength is produced when an electron falls from infinitely far away down to the 2nd level. Using the given constant, this wavelength is exactly \( 4000\text{ \AA} \).

Exam Tip: Remember that "shortest wavelength" always corresponds to the transition starting from \( n = \infty \).

Question 825. Calculate the wavelength of radiation emitted when electron in a hydrogen atom jumps from \( \infty \) to \( n = 1 \). 
Answer:
The transition is from \( n_2 = \infty \) to \( n_1 = 1 \). Using the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R (1 - 0) = R \]
\( \implies \lambda = \frac{1}{R} \)
Using the standard value of Rydberg's constant \( R = 1.097 \times 10^7\text{ m}^{-1} \): \[ \lambda = \frac{1}{1.097 \times 10^7\text{ m}^{-1}} \approx 9.11 \times 10^{-8}\text{ m} = 911\text{ \AA} \]
Thus, the wavelength of the emitted radiation is **\( 911\text{ \AA} \)**.
In simple words: When an electron falls from completely outside the atom straight down to the lowest orbit, it emits ultraviolet light with a wavelength of \( 911\text{ \AA} \).

Exam Tip: Show that \( \frac{1}{\infty^2} = 0 \) to make your step-by-step math complete.

Question 826. (i) Write the relation between mass number and radius of a nucleus. 
(ii) Show that nuclear density in a given nucleus is independent of mass number A. 
Answer:
(i) The relation between the radius \( R \) of a nucleus and its mass number \( A \) is given by: \[ R = R_0 A^{1/3} \] where \( R_0 \) is a constant with an approximate value of \( 1.2 \times 10^{-15}\text{ m} \) (or \( 1.2\text{ fm} \)).

(ii) Let \( m \) be the average mass of a nucleon (proton or neutron). The total mass of a nucleus containing \( A \) nucleons is: \[ M = A \cdot m \] The volume \( V \) of a spherical nucleus of radius \( R \) is: \[ V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A \] The nuclear density \( \rho \) is defined as mass divided by volume: \[ \rho = \frac{M}{V} = \frac{A \cdot m}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3} \] Since \( m \) and \( R_0 \) are constants, the nuclear density \( \rho \) is a constant and is independent of the mass number \( A \).
In simple words: (i) The size of a nucleus grows with the cube root of the number of nucleons it contains. (ii) Since both the mass and the volume grow proportionally with the number of particles, the density remains the same for all nuclei.

Exam Tip: In the density derivation, clearly show the cancellation of \( A \) from the numerator and denominator to prove independence.

Question 827. Compare the radii of two nuclei with mass numbers 1 and 27 respectively. 
Answer:
The radius of a nucleus is related to its mass number \( A \) by: \[ R = R_0 A^{1/3} \] For two nuclei with mass numbers \( A_1 = 1 \) and \( A_2 = 27 \), the ratio of their radii is: \[ \frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{27}\right)^{1/3} = \frac{1}{3} \]
Thus, the ratio of their radii is **1 : 3**.
In simple words: Since the nuclear radius is proportional to the cube root of the mass number, a nucleus with 27 particles has 3 times the radius of a nucleus with 1 particle.

Exam Tip: Write down the cube root relation before plugging in the values to establish the method.

Question 828. What is the nuclear radius of \( ^{125}Fe \), if that of \( ^{27}Al \) is 3.6 Fermi ? 
Answer:
Let \( A_1 = 27 \) and \( R_1 = 3.6\text{ fm} \) be the mass number and radius of the Aluminum nucleus, and \( A_2 = 125 \) be the mass number of the Iron isotope. The relation between nuclear radius and mass number is: \[ R \propto A^{1/3} \] Taking the ratio of the two radii: \[ \frac{R_2}{R_1} = \left(\frac{A_2}{A_1}\right)^{1/3} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3} \]
\( \implies R_2 = R_1 \times \frac{5}{3} \)
Substituting \( R_1 = 3.6\text{ fm} \): \[ R_2 = 3.6 \times \frac{5}{3} = 1.2 \times 5 = 6.0\text{ fm} \]
Thus, the nuclear radius is **6.0 Fermi**.
In simple words: Since the radius scales with the cube root of the mass number, the ratio of the radii is 5 to 3. This makes the radius of the heavier nucleus \( 6\text{ Fermi} \).

Exam Tip: Leave the final answer in Fermi (fm) since the given value was in Fermi, avoiding unnecessary conversions.

Question 829. Two nuclei have mass numbers in the ratio 1:2. What is the ratio of their nuclear densities ? 
Answer:
The nuclear density of any nucleus is independent of its mass number and is given by the constant: \[ \rho = \frac{3m}{4\pi R_0^3} \] Since the density does not depend on the number of nucleons \( A \), both nuclei have the same density. Therefore, the ratio of their nuclear densities is **1 : 1**.
In simple words: All atomic nuclei are made of the same tightly packed material, so they all have the exact same density, giving a ratio of 1 to 1.

Exam Tip: State the physical reason (nuclear density is constant and independent of \( A \)) to justify the 1:1 ratio.

Question 830. What are nuclear forces ? State any two characteristic properties of nuclear forces. 
Answer:
**Nuclear Forces:**
Nuclear forces are the extremely strong, short-range attractive forces that bind protons and neutrons (nucleons) together inside the atomic nucleus, overcoming the strong electrostatic repulsion between the positively charged protons.

**Key Properties (any two):**
1. **Short-range:** These forces operate over extremely tiny distances, typically up to only \( 2-3\text{ fm} \). Beyond this distance, the force drops rapidly to zero.
2. **Charge-independent:** The nuclear force between a proton-proton pair, a neutron-neutron pair, and a proton-neutron pair is virtually identical.
3. **Non-central forces:** Unlike gravitational or electrostatic forces, nuclear forces do not act solely along the straight line joining the centers of the nucleons.
4. **Non-inverse square law:** They do not follow the inverse-square law of distance.
In simple words: Nuclear forces are the powerful, short-distance "glue" that holds protons and neutrons together inside a nucleus. They work equally well regardless of charge and only act over very short ranges.

Exam Tip: Stating "short-range" and "charge-independent" is usually the easiest and most standard combination for full points.

Question 831. Define the term mass defect. 
Answer:
Mass defect (\( \Delta m \)) is defined as the difference between the sum of the individual masses of the constituent nucleons (protons and neutrons) when they are separated, and the actual mass of the stable nucleus they form. The formula is: \[ \Delta m = [Z \cdot m_p + (A-Z) \cdot m_n] - M \] where \( Z \) is the atomic number, \( A \) is the mass number, \( m_p \) is the mass of a proton, \( m_n \) is the mass of a neutron, and \( M \) is the measured mass of the nucleus.
In simple words: When protons and neutrons combine to form a nucleus, a tiny fraction of their mass is converted into binding energy. The missing mass is called the mass defect.

Exam Tip: Be sure to write the formula along with the verbal definition to ensure complete credit.

Question 832. Define binding energy of a nucleus. 
Answer:
The binding energy of a nucleus is defined as the minimum energy required to completely separate its constituent nucleons (protons and neutrons) from each other and place them at rest at an infinite distance apart.
It is also equivalent to the energy released when the individual nucleons combine to form the nucleus, and is calculated from the mass defect (\( \Delta m \)) using Einstein's mass-energy relation: \[ E_b = \Delta m \cdot c^2 \]
In simple words: It is the total energy required to break a nucleus apart into its individual protons and neutrons.

Exam Tip: Emphasize that binding energy is the energy equivalent of the mass defect to show a complete understanding of the concept.

Question 833. What is meant by the term binding energy per nucleon 
Answer:
Binding energy per nucleon (\( E_{bn} \)) is the average energy required to remove a single nucleon (either a proton or a neutron) from the nucleus. It is calculated by dividing the total binding energy \( E_b \) of the nucleus by its mass number \( A \): \[ E_{bn} = \frac{E_b}{A} = \frac{\Delta m \cdot c^2}{A} \] This value serves as a direct measure of the thermodynamic stability of the nucleus.
In simple words: It is the total nuclear binding energy divided by the number of particles in the nucleus, showing how tightly each individual particle is held.

Exam Tip: Always mention that binding energy per nucleon determines the overall stability of a nucleus, as this is a key conceptual point.

Question 834. The binding energies of deuteron (\( _1^2H \)) and \( \alpha \)-particle (\( _2^4He \)) are 1.25 and 7.2 MeV/ nucleon respectively. Which nucleus is more stable ?
Answer:
The relative stability of a nucleus is determined by its binding energy per nucleon, not by its total binding energy. Here: - Binding energy per nucleon of deuteron (\( _1^2H \)) = \( 1.25\text{ MeV} \) - Binding energy per nucleon of \( \alpha \)-particle (\( _2^4He \)) = \( 7.2\text{ MeV} \)
Since the binding energy per nucleon of the \( \alpha \)-particle is significantly higher, it is much more stable than the deuteron.
In simple words: The alpha particle is much more stable because its nucleons are bound together with several times more energy on average than those in a deuteron.

Exam Tip: Clearly state that stability depends on "binding energy per nucleon" to ensure you get full credit for your reasoning.

Question 835. Which out of two nuclei \( _3^7Li \) & \( _4^7Be \) is more stable ?
Answer:
The Lithium nucleus \( _3^7Li \) is more stable than the Beryllium nucleus \( _4^7Be \).

**Reasoning:**
For \( _3^7Li \), the number of protons is \( Z = 3 \), and the number of neutrons is \( N = A - Z = 7 - 3 = 4 \). The neutron-to-proton ratio is: \[ \frac{N}{Z} = \frac{4}{3} \approx 1.33 \]
For \( _4^7Be \), the number of protons is \( Z = 4 \), and the number of neutrons is \( N = 7 - 4 = 3 \). The neutron-to-proton ratio is: \[ \frac{N}{Z} = \frac{3}{4} = 0.75 \]
A nucleus is generally more stable if it has a higher neutron-to-proton ratio to help offset the electrostatic repulsion between protons, or a higher binding energy per nucleon.
In simple words: Lithium-7 is more stable because it has a better balance of neutrons to protons, which helps screen the repulsive forces between the positive charges.

Exam Tip: Mention the \( N/Z \) ratio or binding energy comparison to provide a complete scientific justification.

Question 836. Why is mass of a nucleus is always less than the sum of the masses of its constituent, neutrons & protons ?
Answer:
When individual protons and neutrons approach each other to form a bound nucleus, they experience strong attractive nuclear forces. This causes their potential energy to decrease significantly, becoming negative. According to Einstein's mass-energy equivalence principle (\( E = mc^2 \)), this reduction in potential energy is accompanied by a corresponding loss in the total mass of the system. This missing mass is released as energy during the formation of the nucleus.
In simple words: When nucleons bind together, some of their mass is converted into energy to hold the nucleus together. This energy release results in a slightly lighter nucleus.

Exam Tip: Reference Einstein's mass-energy relation (\( \Delta E = \Delta m \cdot c^2 \)) to explain how the potential energy drop leads to the mass difference.

Question 837. If the nucleons of a nucleus are separated far apart from each other, the sum of the masses of all these nucleons is larger than the mass of the nucleus. Why ? 
Answer:
To pull the bound nucleons apart to an infinite separation, work must be done against the strong attractive nuclear forces. This requires supplying an external energy equal to the binding energy of the nucleus. According to Einstein's mass-energy relation (\( E = mc^2 \)), this added energy is converted back into mass, which increases the total mass of the individual, separated nucleons.
In simple words: You must add energy to break the nuclear bonds. This added energy turns back into mass, making the separated particles heavier than the original bound nucleus.

Exam Tip: Explain the process from the perspective of work done against attractive forces to make your conceptual answer stand out.

Question 838. If the total number of neutrons & protons in a nuclear reaction is conserved, how is then the energy is absorbed or evolved in the reaction ? 
OR
In a nuclear reaction,
\( _1^2H + _1^2H \to _2^3He + _0^1n + 3.27\text{ MeV} \)
though the number of nucleons is conserved on both sides of the reaction, yet the energy is released. How ? Explain.
Answer:
Although the total number of nucleons (protons and neutrons) is conserved in a nuclear reaction, the total rest mass of the reactant nuclei is not equal to the total rest mass of the product nuclei. During the reaction, the nucleons rearrange themselves into configurations with higher binding energy per nucleon. This increase in binding energy corresponds to a decrease in the total rest mass of the system (mass defect). The lost mass \( \Delta m \) is converted into energy and released according to Einstein's mass-energy equivalence: \[ E = \Delta m \cdot c^2 \]
In simple words: Even though the total number of particles stays the same, the product particles are bound together more tightly and are slightly lighter. The missing mass is converted into the energy released in the reaction.

Exam Tip: Emphasize that it is the "rest mass" or "binding energy per nucleon" that changes during the rearrangement, even though the nucleon count remains conserved.

Question 839. A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks in to two fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the energy released.
Answer:
The total binding energy of the parent nucleus (\( A = 240 \)) is: \[ E_{i} = 240 \times 7.6\text{ MeV} = 1824\text{ MeV} \]
When it fissions into two fragments of \( A = 120 \) each, the total binding energy of the two product fragments is: \[ E_{f} = 2 \times (120 \times 8.5\text{ MeV}) = 240 \times 8.5\text{ MeV} = 2040\text{ MeV} \]
The energy released (\( Q \)) in this fission process is the difference in binding energies: \[ Q = E_{f} - E_{i} = 2040\text{ MeV} - 1824\text{ MeV} = 216\text{ MeV} \]
Thus, the total energy released is **\( 216\text{ MeV} \)**.
In simple words: The products are bound more tightly than the parent nucleus, with an increase in binding energy from \( 1824\text{ MeV} \) to \( 2040\text{ MeV} \). This difference releases \( 216\text{ MeV} \) of energy.

Exam Tip: Double-check your arithmetic when calculating \( 240 \times (8.5 - 7.6) \), as this is a very common place for quick mental math errors.

Question 840. Calculate the energy released in the fusion reaction : 
\( _1^2H + _1^2H \to _2^3He + _0^1n \), where BE of \( _1^2H = 2.23\text{ MeV} \) and of \( _2^3He = 7.73\text{ MeV} \)
Answer:
The total binding energy of the reactants (two Deuteron nuclei) is: \[ E_{i} = 2 \times 2.23\text{ MeV} = 4.46\text{ MeV} \]
The total binding energy of the products (Helium-3 and a free neutron, which has zero binding energy) is: \[ E_{f} = 7.73\text{ MeV} + 0 = 7.73\text{ MeV} \]
The energy released (\( Q \)) during this fusion reaction is: \[ Q = E_{f} - E_{i} = 7.73\text{ MeV} - 4.46\text{ MeV} = 3.27\text{ MeV} \]
Thus, the energy released is **\( 3.27\text{ MeV} \)**.
In simple words: The starting materials have a binding energy of \( 4.46\text{ MeV} \), while the final products are bound with \( 7.73\text{ MeV} \). The difference of \( 3.27\text{ MeV} \) is released as energy.

Exam Tip: Remember that free individual nucleons (like the neutron on the right side) have a binding energy of zero.

Question 841. The energy levels of a hypothetical atom are shown below. Which of the shown transitions will result in the emission of photon of wavelength 275 nm ? 
Answer:
The wavelength of the emitted photon is \( \lambda = 275\text{ nm} = 275 \times 10^{-9}\text{ m} \). The energy difference \( \Delta E \) corresponding to this wavelength is: \[ \Delta E = \frac{hc}{\lambda} \] Substituting the standard values (\( h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} \), \( c = 3 \times 10^8\text{ m/s} \)): \[ \Delta E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{275 \times 10^{-9}}\text{ J} \] \[ \Delta E = \frac{1.98 \times 10^{-25}}{2.75 \times 10^{-7}}\text{ J} = 7.2 \times 10^{-19}\text{ J} \] Converting this energy value into electron-volts: \[ \Delta E = \frac{7.2 \times 10^{-19}}{1.6 \times 10^{-19}}\text{ eV} = 4.5\text{ eV} \]
Now, checking the energy differences for the transitions shown in the diagram: - Transition A: from \( -2\text{ eV} \) to \( 0\text{ eV} \), \( \Delta E = 2\text{ eV} \) - Transition B: from \( -4.5\text{ eV} \) to \( 0\text{ eV} \), \( \Delta E = 4.5\text{ eV} \) - Transition C: from \( -4.5\text{ eV} \) to \( -2\text{ eV} \), \( \Delta E = 2.5\text{ eV} \) - Transition D: from \( -10\text{ eV} \) to \( -2\text{ eV} \), \( \Delta E = 8\text{ eV} \)
Since the calculated energy difference of \( 4.5\text{ eV} \) matches Transition B, the correct transition is **B**.
In simple words: The energy of a \( 275\text{ nm} \) photon is exactly \( 4.5\text{ eV} \). Looking at the energy levels, only transition B covers this exact energy jump.

Exam Tip: Clearly show the conversion from Joules to eV to justify your choice of transition state.

Question 842. Calculate the binding energy per nucleon of \( _{20}^{40}Ca \) nucleus. (Given, Mass of \( _{20}^{40}Ca \) = 39.962589 u, Mass of proton= 1.007825 u, Mass of neutron =1.008665 u & 1u= 931 MeV/\(c^2\)) 
Answer:
For the Calcium nucleus \( _{20}^{40}Ca \), the number of protons is \( Z = 20 \) and the number of neutrons is \( N = A - Z = 40 - 20 = 20 \).
The mass of 20 free protons is: \[ 20 \times 1.007825\text{ u} = 20.156500\text{ u} \] The mass of 20 free neutrons is: \[ 20 \times 1.008665\text{ u} = 20.173300\text{ u} \]
The total mass of the individual, separated constituents is: \[ 20.156500 + 20.173300 = 40.329800\text{ u} \]
The mass defect \( \Delta m \) is the difference between this sum and the measured nuclear mass: \[ \Delta m = 40.329800\text{ u} - 39.962589\text{ u} = 0.367211\text{ u} \]
The total binding energy \( E_b \) of the nucleus is: \[ E_b = 0.367211 \times 931\text{ MeV} \approx 341.87\text{ MeV} \]
The binding energy per nucleon is: \[ E_{bn} = \frac{E_b}{A} = \frac{341.87\text{ MeV}}{40} \approx 8.547\text{ MeV/nucleon} \]
Thus, the binding energy per nucleon of the calcium nucleus is **\( 8.547\text{ MeV/nucleon} \)**.
In simple words: The mass defect of the Calcium nucleus is \( 0.367211\text{ u} \), which corresponds to a total binding energy of \( 341.87\text{ MeV} \). Dividing this by the 40 nucleons gives \( 8.547\text{ MeV} \) of binding energy per nucleon.

Exam Tip: Pay close attention to decimal places during mass defect calculations, as rounding errors can significantly affect the final binding energy value.

Question 843. What is radioactivity ? 
Answer:
Radioactivity is the spontaneous and continuous disintegration of unstable atomic nuclei, accompanied by the emission of ionizing particles (such as alpha or beta particles) and high-energy electromagnetic radiation (gamma rays) to achieve a more stable nuclear configuration.
In simple words: It is the natural process where unstable heavy atoms break down on their own, releasing radiation in the form of particles or rays to become stable.

Exam Tip: Use keywords like "spontaneous" and "continuous" to describe the decay process accurately.

Question 844. When a radioactive radiation is placed in an electric or magnetic field it divides in to three parts. Why ? 
Answer:
A radioactive substance can emit three distinct types of radiation: alpha (\( \alpha \)) particles, beta (\( \beta \)) particles, and gamma (\( \gamma \)) rays. When these radiations pass through an electric or magnetic field, they behave differently because of their electrical charges: 1. **Alpha particles** carry a positive charge (\( +2e \)) and are deflected toward the negative plate or in accordance with the magnetic force direction. 2. **Beta particles** carry a negative charge (\( -e \)) and are deflected in the opposite direction (toward the positive plate). 3. **Gamma rays** are uncharged, neutral high-energy photons, so they pass straight through the field without any deflection.
In simple words: The radiation splits into three paths because it contains positively charged alpha particles, negatively charged beta particles, and neutral gamma rays, which react differently to the field.

Exam Tip: Briefly state the charge of each of the three types of radiation to explain their deflection behaviors clearly.

Question 845. Why do \alpha -particles have high ionising power ? 
Answer:
Alpha particles are Helium nuclei (\( _2^4He \)), which have a relatively large rest mass and a double positive charge (\( +2e \)). Because of this high mass and charge, they move relatively slowly and interact strongly with the electrons of the atoms they pass through, easily stripping them away and resulting in high ionizing power.
In simple words: Alpha particles are heavy and carry a strong positive charge, which makes it easy for them to pull electrons off the atoms they bump into.

Exam Tip: Mention both the large mass and the double positive charge to explain the high ionization power completely.

Question 846. Which of the following radiations \alpha -rays, \beta -rays, \gamma -rays 
(i) are similar to X-rays
(ii) are easily absorbed by the matter
(iii) travel with greatest speed
(iv) are similar in nature to cathode rays.
Answer:
(i) **Similar to X-rays:** Gamma (\( \gamma \)) rays, as both are high-frequency electromagnetic waves.

(ii) **Easily absorbed by matter:** Alpha (\( \alpha \)) rays, because of their large mass, low speed, and high charge, which causes them to interact strongly and lose energy quickly.

(iii) **Travel with greatest speed:** Gamma (\( \gamma \)) rays, which travel at the speed of light in vacuum (\( 3 \times 10^8\text{ m/s} \)).

(iv) **Similar in nature to cathode rays:** Beta (\( \beta^- \)) rays, as both consist of high-speed streams of electrons.
In simple words: (i) Gamma rays are like X-rays. (ii) Alpha rays are easily stopped by paper or skin. (iii) Gamma rays travel the fastest. (iv) Beta rays are streams of fast electrons, just like cathode rays.

Exam Tip: Make sure you address all four sub-parts systematically using the matching names to prevent losing presentation marks.

Question 847. What is the difference between an electron and a \beta - particle ? 
Answer:
Physically, an electron and a beta particle are identical, sharing the same mass, charge, and spin. The only difference is their origin: an ordinary electron resides in the atomic shells orbiting the nucleus, whereas a beta particle is created and ejected from within the nucleus during the radioactive decay of a neutron into a proton.
In simple words: They are the exact same particle, but an electron orbits on the outside of the atom, while a beta particle is shot out from deep inside the nucleus.

Exam Tip: Focus on the "origin" of the particles as the primary scientific difference between them.

Question 848. A nucleus contains no electrons, yet it ejects them. How ? 
Answer:
Although there are no electrons inside a nucleus, a beta electron (\( \beta^- \)) is created at the instant of radioactive decay. Inside an unstable nucleus, a neutron spontaneously converts into a proton, an electron, and an antineutrino: \[ _0^1n \to _1^1p + _{-1}^0e + \bar{\nu} \] The newly created proton remains within the nucleus, while the high-energy electron (\( \beta^- \)) and the antineutrino are immediately ejected.
In simple words: The nucleus does not store electrons. Instead, it creates one on the spot when a neutron decays into a proton, releasing the electron as a beta particle.

Exam Tip: Write the basic decay equation \( n \to p + e^- + \bar{\nu} \) to back up your verbal explanation.

Question 849. A nucleus undergoes \beta ^- -decay. How does its (i) mass number (ii) atomic number change ?
Answer:
During \( \beta^- \) decay, a neutron inside the nucleus converts into a proton and an electron. 1. **Mass Number (\( A \)):** The total number of nucleons (protons + neutrons) remains **unchanged**, because one neutron is replaced by one proton. 2. **Atomic Number (\( Z \)):** The atomic number **increases by one** (\( Z' = Z + 1 \)), because the nucleus now contains one additional proton.
In simple words: (i) The total weight (mass number) stays the same because the total count of nuclear particles is unchanged. (ii) The atomic number goes up by one because a neutron turned into a proton.

Exam Tip: State clearly that the total count of nucleons remains constant, which is why the mass number \( A \) does not change.

Question 850. What is \beta -decay ?
Answer:
Beta (\( \beta \)) decay is a spontaneous nuclear transition in which an unstable nucleus achieves a more stable configuration by emitting a beta particle (either a high-speed electron \( \beta^- \) or a positron \( \beta^+ \)), along with an antineutrino or neutrino.
In simple words: It is a type of radioactivity where a nucleus shoots out a fast electron or positron to become more stable.

Exam Tip: Mention both \( \beta^- \) (electron) and \( \beta^+ \) (positron) emissions to show a complete understanding of beta decay.

Question 851. (i) Write the nuclear decay process of \beta ^- -decay \( _{15}^{32}P \).
(ii) Write the \beta ^- -decay of tritium in symbolic form. 
Answer:
(i) The \( \beta^- \) decay of Phosphorus-32 is represented by: \[ _{15}^{32}P \to _{16}^{32}S + _{-1}^0e + \bar{\nu} \]
(ii) Tritium (\( _1^3H \)) is an isotope of hydrogen. Its \( \beta^- \) decay process is: \[ _1^3H \to _2^3He + _{-1}^0e + \bar{\nu} \] where \( \bar{\nu} \) represents the antineutrino emitted to conserve lepton number and spin.
In simple words: (i) Phosphorus-32 decays into Sulfur-32 by emitting a beta electron and an antineutrino. (ii) Tritium decays into Helium-3 in the same way.

Exam Tip: Do not forget to include the antineutrino (\( \bar{\nu} \)) in the decay equations, as it is required for physical conservation laws.

Question 852. Write the basic nuclear process involved in the emission of (a) \beta ^- decay and (b) \beta ^+ decay in a symbolic form, by a radioactive nucleus. 
Answer:
The basic nuclear processes occurring at the nucleon level during beta decays are:

(a) **For \( \beta^- \) decay (neutron-to-proton conversion):** \[ _0^1n \to _1^1p + _{-1}^0e + \bar{\nu} \] A neutron transforms into a proton, emitting an electron (\( \beta^- \)) and an antineutrino (\( \bar{\nu} \)).

(b) **For \( \beta^+ \) decay (proton-to-neutron conversion):** \[ _1^1p \to _0^1n + _{+1}^0e + \nu \] A proton transforms into a neutron, emitting a positron (\( \beta^+ \)) and a neutrino (\( \nu \)).
In simple words: (a) In \( \beta^- \) decay, a neutron turns into a proton, releasing an electron and an antineutrino. (b) In \( \beta^+ \) decay, a proton turns into a neutron, releasing a positron and a neutrino.

Exam Tip: Pair \( \beta^- \) with the antineutrino (\( \bar{\nu} \)) and \( \beta^+ \) with the neutrino (\( \nu \)) to show correct lepton conservation.

Question 853. Why is the detection of neutrinos found very difficult ? 
Answer:
Neutrinos are elementary particles that carry no electrical charge and have an extremely tiny, nearly zero rest mass. Because they are neutral and do not feel the electromagnetic force, they interact only via the extremely weak force and gravity. This allows them to pass through vast amounts of solid matter (like the Earth) without undergoing any collisions, making their detection exceptionally difficult.
In simple words: Neutrinos have no charge, almost no mass, and rarely interact with matter, allowing them to pass straight through stars and planets undetected.

Exam Tip: Use terms like "neutral charge", "nearly massless", and "weak interaction with matter" in your explanation.

Question 854. Why the mass number of a nuclide undergoing \beta -decay does not change? 
OR
In both \beta -decay process, the mass number of the nucleus remains same, whereas the atomic number Z increases by one in \beta ^- decay and decreases by one in \beta ^+ decay. Explain giving reason. 
Answer:
During any beta decay process, the nuclear change involves the conversion of one type of nucleon into another. Specifically: - In \( \beta^- \) decay, a neutron inside the nucleus transforms into a proton:
\( _0^1n \to _1^1p + _{-1}^0e + \bar{\nu} \)
- In \( \beta^+ \) decay, a proton inside the nucleus transforms into a neutron:
\( _1^1p \to _0^1n + _{+1}^0e + \nu \)

In both cases, the total count of nucleons (\( A = Z + N \)) remains constant, because the loss of one nucleon of one type is balanced by the gain of one nucleon of the other type. Since protons and neutrons have nearly identical masses, the mass number \( A \) does not change.
However, the atomic number \( Z \) is the proton count. It increases by one in \( \beta^- \) decay (due to the new proton) and decreases by one in \( \beta^+ \) decay (due to the loss of a proton).
In simple words: The mass number doesn't change because a neutron turns into a proton (or vice versa), keeping the total number of heavy particles constant. The atomic number changes because the number of protons increases or decreases by one.

Exam Tip: Write down both the \( \beta^- \) and \( \beta^+ \) processes to support your explanation of why the total nucleon count remains conserved.

Question 855. Write the nuclear reactions for the following- 
(i) \alpha -decay of \( _{94}^{242}Pu \)
(ii) \beta ^- -decay of \( _{15}^{32}P \)
(iii) \beta ^+ decay of \( _6^{11}C \)
(iv) \alpha -decay of \( _{88}^{226}Ra \)
Answer:
(i) **\( \alpha \)-decay of Plutonium-242:**
An alpha decay reduces the mass number by 4 and the atomic number by 2: \[ _{94}^{242}Pu \to _{92}^{238}U + _2^4He \]
(ii) **\( \beta^- \)-decay of Phosphorus-32:**
This increases the atomic number by 1: \[ _{15}^{32}P \to _{16}^{32}S + _{-1}^0e + \bar{\nu} \]
(iii) **\( \beta^+ \)-decay of Carbon-11:**
This decreases the atomic number by 1: \[ _6^{11}C \to _5^{11}B + _{+1}^0e + \nu \]
(iv) **\( \alpha \)-decay of Radium-226:**
This reduces the mass number by 4 and the atomic number by 2: \[ _{88}^{226}Ra \to _{86}^{222}Rn + _2^4He \]
In simple words: (i) Plutonium decays to Uranium and an alpha particle. (ii) Phosphorus decays to Sulfur and a beta electron. (iii) Carbon decays to Boron and a positron. (iv) Radium decays to Radon and an alpha particle.

Exam Tip: Always make sure both the mass numbers (top) and the atomic numbers (bottom) balance perfectly on both sides of each equation.

Question 856. In the reactions given below, find the values of x, y & z and a, b & c. 
(a) \( _6^{11}C \to _y^{z}B + x + \nu \)
(b) \( _6^{12}C + _6^{12}C \to _{a}^{20}Ne + _b^{c}He \)
Answer:
(a) The given reaction is a \( \beta^+ \) decay: \[ _6^{11}C \to _y^{z}B + x + \nu \] Balancing the atomic numbers (bottom): \[ 6 = y + Z_x + 0 \] Since \( \nu \) is a neutrino, \( x \) must be a positron (\( _{+1}^0e \)). Thus, \( Z_x = 1 \): \[ 6 = y + 1 \implies y = 5 \] Balancing the mass numbers (top): \[ 11 = z + 0 \implies z = 11 \] Thus, the values are: \[ x = _{+1}^0e \quad (\text{or } e^+), \quad y = 5, \quad z = 11 \]
(b) The fusion reaction is: \[ _6^{12}C + _6^{12}C \to _{a}^{20}Ne + _b^{c}He \] Balancing the mass numbers (top): \[ 12 + 12 = 20 + c \implies 24 = 20 + c \implies c = 4 \] Balancing the atomic numbers (bottom): The product is a Helium nucleus (\( _2^4He \)), so \( b = 2 \): \[ 6 + 6 = a + b \implies 12 = a + 2 \implies a = 10 \] Thus, the values are: \[ a = 10, \quad b = 2, \quad c = 4 \]
In simple words: (a) In the carbon decay, \( x \) is a positron, B has an atomic number of 5 and a mass of 11. (b) In the carbon fusion, Ne has an atomic number of 10, and Helium has an atomic number of 2 and a mass of 4.

Exam Tip: Write down the conservation equations for both mass number and atomic number to make your working clear to the examiner.

Question 857. In the following nuclear reaction assign the value of Z and A. 
\( _0^1n + _{92}^{235}U \to _{56}^{144}Ba + _{Z}^{A}Kr + 3_0^1n \)
Answer:
The nuclear reaction represents the fission of Uranium-235: \[ _0^1n + _{92}^{235}U \to _{56}^{144}Ba + _{Z}^{A}Kr + 3(_0^1n) \]
Balancing the mass numbers (top) on both sides: \[ 1 + 235 = 144 + A + 3(1) \] \[ 236 = 147 + A \implies A = 236 - 147 = 89 \]
Balancing the atomic numbers (bottom) on both sides: \[ 0 + 92 = 56 + Z + 3(0) \] \[ 92 = 56 + Z \implies Z = 92 - 56 = 36 \]
Thus, the values are **\( Z = 36 \)** and **\( A = 89 \)** (the isotope is \( _{36}^{89}Kr \)).
In simple words: By making sure the total mass and atomic numbers are the same before and after the fission, we find that Krypton has a mass number of 89 and an atomic number of 36.

Exam Tip: Remember to multiply the nucleon numbers of the neutrons by 3, as there are three individual neutrons produced in the reaction.

Question 858. Identify the nature of the radioactive radiations emitted in each step of the decay process given below: 
\( _Z^AX \to _{Z-2}^{A-4}Y \to _{Z-1}^{A-4}W \to _{Z-1}^{A-4}W \)
Answer:
Let us analyze each transition step-by-step:

1. **First Step (\( _Z^AX \to _{Z-2}^{A-4}Y \)):**
The mass number decreases by 4, and the atomic number decreases by 2. This represents the emission of an **alpha (\( \alpha \)) particle** (\( _2^4He \)).

2. **Second Step (\( _{Z-2}^{A-4}Y \to _{Z-1}^{A-4}W \)):**
The mass number remains unchanged, while the atomic number increases by 1. This represents the emission of a **beta (\( \beta^- \)) particle** (electron, \( _{-1}^0e \)).

3. **Third Step (\( _{Z-1}^{A-4}W \to _{Z-1}^{A-4}W \)):**
Both the mass number and the atomic number remain completely unchanged, indicating a transition from an excited state to a lower energy state. This represents the emission of a **gamma (\( \gamma \)) ray** (high-energy photon).
In simple words: The first step emits an alpha particle (losing mass and charge). The second step emits a beta electron (gaining charge). The third step emits a gamma ray (no change in mass or charge).

Exam Tip: List the three emitted radiations in the correct chronological order (\( \alpha \), then \( \beta^- \), then \( \gamma \)) to answer the question clearly.

Question 859. Give the mass number and atomic number of elements on the right hand side of the decay process. 
\( _{86}^{220}Ru \to Po + He \)
Answer:
The reaction represents an alpha decay process: \[ _{86}^{220}Rn \to _{Z}^{A}Po + _2^4He \] *(Note: The typo 'Ru' in the question is corrected to 'Rn' for Radon, and 'He' represents the alpha particle).*

Balancing the mass numbers (top): \[ 220 = A + 4 \implies A = 216 \]
Balancing the atomic numbers (bottom): \[ 86 = Z + 2 \implies Z = 84 \]
Thus, the completed nuclear reaction is: \[ _{86}^{220}Rn \to _{84}^{216}Po + _2^4He \] The products on the right hand side are:
- **Polonium (Po):** Mass number \( A = 216 \), Atomic number \( Z = 84 \)
- **Helium (He):** Mass number \( A = 4 \), Atomic number \( Z = 2 \)
In simple words: After Radium decays, the Polonium has an atomic number of 84 and a mass of 216, while the Helium has an atomic number of 2 and a mass of 4.

Exam Tip: Always make sure to write the full final balanced reaction equation to clarify your calculations.

Question 860. A radioactive nucleus ‘A’ undergoes series of decays shown in the following scheme : 
\( A \xrightarrow{\alpha} A_1 \xrightarrow{\beta} A_2 \xrightarrow{\gamma} A_3 \)
If mass number and atomic number of \( A_3 \) are 176 and 69 respectively, find the mass number and atomic number of A
Answer:
Let us trace the decay series in reverse from \( A_3 \) to the parent nucleus \( A \):
We are given that \( A_3 = _{69}^{176}A_3 \).

1. **From \( A_2 \to A_3 \) via \( \gamma \) emission:**
Gamma decay does not alter the mass number or the atomic number of the nucleus. Thus, the parent \( A_2 \) is identical to \( A_3 \): \[ A_2 = _{69}^{176}A_2 \]

2. **From \( A_1 \to A_2 \) via \( \beta^- \) emission:**
Beta-minus decay increases the atomic number by 1 while keeping the mass number the same. Therefore, the parent nucleus \( A_1 \) must have an atomic number that is one less: \[ Z_{A1} = 69 - 1 = 68, \quad A_{A1} = 176 \] \[ A_1 = _{68}^{176}A_1 \]

3. **From \( A \to A_1 \) via \( \alpha \) emission:**
Alpha decay reduces the mass number by 4 and the atomic number by 2. Therefore, the original parent nucleus \( A \) must have: \[ Z_A = 68 + 2 = 70, \quad A_A = 176 + 4 = 180 \] \[ A = _{70}^{180}A \]
Thus, the mass number of \( A \) is **180** and its atomic number is **70**.
In simple words: Working backward from \( A_3 \) (176, 69): the gamma decay means \( A_2 \) is also (176, 69). The beta decay means \( A_1 \) was (176, 68). The alpha decay means the original nucleus \( A \) was (180, 70).

Exam Tip: Be careful when working backward; remember that moving backward through an alpha decay requires adding 4 to the mass number and 2 to the atomic number.

Question 861. A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme- 
\( A \xrightarrow{\alpha} A_1 \xrightarrow{\beta} A_2 \xrightarrow{\alpha} A_3 \xrightarrow{\gamma} A_4 \)
If the mass number and atomic number of A are 180 & 72 respectively, What are these numbers for \( A_4 \) ?
Answer:
We start with the parent nucleus \( A = _{72}^{180}A \). Let us trace the decay steps chronologically:

1. **First Step: \( A \xrightarrow{\alpha} A_1 \)**
Alpha decay reduces the mass number by 4 and the atomic number by 2: \[ A_1 = _{72-2}^{180-4}A_1 = _{70}^{176}A_1 \]

2. **Second Step: \( A_1 \xrightarrow{\beta^-} A_2 \)**
Beta-minus decay increases the atomic number by 1 while keeping the mass number the same: \[ A_2 = _{70+1}^{176}A_2 = _{71}^{176}A_2 \]

3. **Third Step: \( A_2 \xrightarrow{\alpha} A_3 \)**
Another alpha decay reduces the mass number by 4 and the atomic number by 2: \[ A_3 = _{71-2}^{176-4}A_3 = _{69}^{172}A_3 \]

4. **Fourth Step: \( A_3 \xrightarrow{\gamma} A_4 \)**
Gamma decay does not change the mass number or the atomic number: \[ A_4 = _{69}^{172}A_4 \]

Thus, for the final nucleus \( A_4 \), the mass number is **172** and the atomic number is **69**.
In simple words: Starting at (180, 72): the first alpha decay gives (176, 70). The beta decay gives (176, 71). The second alpha decay gives (172, 69). The gamma decay leaves it unchanged at (172, 69).

Exam Tip: List each intermediate nucleus (\( A_1, A_2, A_3 \)) with its calculated nucleon values to show a clear, logical progression.

Question 862. A radioactive isotope D decays according to the sequence - 
\( D \xrightarrow{\beta^-} D_1 \xrightarrow{\alpha} D_2 \)
If the mass number & atomic number for \( D_2 \) are 176 & 71 respectively, find the mass number and atomic number of D.
Answer:
We are given the final product \( D_2 = _{71}^{176}D_2 \). Working backward:

1. **From \( D_1 \to D_2 \) via \( \alpha \) emission:**
Alpha decay reduces the mass number by 4 and the atomic number by 2. Therefore, the parent nucleus \( D_1 \) must have: \[ Z_{D1} = 71 + 2 = 73, \quad A_{D1} = 176 + 4 = 180 \] \[ D_1 = _{73}^{180}D_1 \]

2. **From \( D \to D_1 \) via \( \beta^- \) emission:**
Beta-minus decay increases the atomic number by 1, while keeping the mass number the same. Therefore, the original parent nucleus \( D \) must have: \[ Z_D = 73 - 1 = 72, \quad A_D = 180 \] \[ D = _{72}^{180}D \]
Thus, the original nucleus \( D \) has a mass number of **180** and an atomic number of **72**.
In simple words: Working backward from \( D_2 \) (176, 71): the alpha decay means \( D_1 \) was (180, 73). The beta decay means the starting nucleus \( D \) was (180, 72).

Exam Tip: Be sure to distinguish between \( \beta^- \) and \( \beta^+ \) decay, as a \( \beta^- \) decay means the daughter nucleus gained a proton relative to the parent.

Question 863. The sequence of stepwise decays of a radioactive nucleus is -
\( D \xrightarrow{\alpha} D_1 \xrightarrow{\beta^-} D_2 \)
If the atomic number and mass number of \( D_2 \) are 71 & 176 respectively, What are their corresponding values for D ?
Answer:
The final product is \( D_2 = _{71}^{176}D_2 \). We work backward step-by-step to find \( D \):

1. **From \( D_1 \to D_2 \) via \( \beta^- \) emission:**
Beta-minus decay increases the atomic number by 1 while keeping the mass number constant. Therefore, the parent nucleus \( D_1 \) must have had: \[ Z_{D1} = 71 - 1 = 70, \quad A_{D1} = 176 \] \[ D_1 = _{70}^{176}D_1 \]

2. **From \( D \to D_1 \) via \( \alpha \) emission:**
Alpha decay reduces the mass number by 4 and the atomic number by 2. Therefore, the original parent nucleus \( D \) must have had: \[ Z_D = 70 + 2 = 72, \quad A_D = 176 + 4 = 180 \] \[ D = _{72}^{180}D \]
Thus, the original nucleus \( D \) has an atomic number of **72** and a mass number of **180**.
In simple words: Working backward from \( D_2 \) (176, 71): the beta decay means \( D_1 \) was (176, 70). The alpha decay means the original nucleus \( D \) was (180, 72).

Exam Tip: Always state both values (atomic number and mass number) explicitly in your final answer line.

Question 864. (a) Write two important limitations of Rutherford nuclear model of the atom. 
(b) How these were explained in Bohr’s model of hydrogen atom ? 
Answer:
(a) **Limitations of Rutherford's Model:**
1. **Instability of the atom:** According to classical electromagnetic theory, an electron revolving in a circular orbit experiences continuous centripetal acceleration. An accelerating charged particle must continuously radiate electromagnetic energy. As it loses energy, its orbit would shrink, causing it to spiral into the nucleus, making the atom highly unstable.
2. **Origin of line spectra:** Since the orbiting radius of the electron would decrease continuously, it should emit electromagnetic waves of all continuously changing frequencies. Thus, the model predicts a continuous spectrum, which contradicts the observed discrete line spectra of atoms.

(b) **Bohr's Explanation:**
1. **Stable Orbits:** Bohr postulated that electrons can revolve only in specific non-radiating paths called stationary orbits, where their orbital angular momentum is quantized: \[ mvr = n\frac{h}{2\pi} \] As long as an electron remains in one of these stable orbits, it does not radiate any energy, ensuring atomic stability.
2. **Discrete Line Spectra:** Energy is emitted or absorbed only when an electron transitions between two stable orbits. This transition produces a photon of a single, discrete frequency: \[ h\nu = E_2 - E_1 \] which explains the origin of the discrete line spectra rather than a continuous one.
In simple words: (a) Rutherford's model couldn't explain why accelerating electrons don't lose energy and spiral into the nucleus, nor why atoms emit specific lines of light instead of a full rainbow. (b) Bohr solved this by stating that electrons are locked into specific "safe" orbits where they cannot lose energy, and only emit light when jumping between these levels.

Exam Tip: Structure your answer clearly with separate headings for "Limitations" and "Bohr's Explanation" to maximize clarity.

Question 865. How does de-Broglie explain the stationary orbits for revolution of electrons using Bohr’s quantization condition ?
Answer:
According to de-Broglie's hypothesis, a moving electron behaves as a wave. For an electron to exist in a stable circular orbit of radius \( r \), its matter wave must form a standing wave. This is possible only if the total circumference of the orbit is an integral multiple of its de-Broglie wavelength \( \lambda \): \[ 2\pi r = n\lambda \] where \( n = 1, 2, 3, \dots \)
According to de-Broglie's relation, the wavelength of the electron is: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] Substituting this wavelength into the circular standing wave condition: \[ 2\pi r = n\left(\frac{h}{mv}\right) \] Rearranging the terms: \[ mvr = n\frac{h}{2\pi} \] This is Bohr's quantization condition for orbital angular momentum. Thus, de-Broglie provided a physical explanation for the existence of stationary orbits as circular standing waves of electron matter.
In simple words: An electron orbiting a nucleus forms a circular wave. For the wave to be stable without destroying itself, the orbit's circumference must fit a whole number of wavelengths. This requirement mathematically leads to Bohr's quantization rule.

Exam Tip: Clearly state the circular standing wave condition \( 2\pi r = n\lambda \) as the starting point of your derivation.

Question 866. Derive the Bohr’s quantization condition for angular momentum of the orbiting of electron in hydrogen atom, Using de-Broglie’s hypothesis. 
Answer:
According to de-Broglie, an electron orbiting in a stable circular path of radius \( r_n \) forms a standing wave. For the wave to constructively interfere and survive, the circumference of the orbit must contain an integral number of wavelengths \( \lambda \): \[ 2\pi r_n = n\lambda \quad \text{for } n = 1, 2, 3, \dots \] According to the de-Broglie relation, the matter wavelength is: \[ \lambda = \frac{h}{mv_n} \] Substituting the value of \( \lambda \) into the standing wave condition: \[ 2\pi r_n = n\left(\frac{h}{mv_n}\right) \] Rearranging this equation to solve for the angular momentum (\( L = m v_n r_n \)): \[ m v_n r_n = n\frac{h}{2\pi} \] This derives Bohr's quantization condition, showing that the orbital angular momentum of the electron is restricted to integral multiples of \( \frac{h}{2\pi} \).
In simple words: By treating the orbiting electron as a wave that must fit perfectly around the orbit's circumference, we derive Bohr's rule that angular momentum must be a multiple of \( h/2\pi \).

Exam Tip: Be sure to state that only integer multiples of wavelength form stable standing waves, which explains why other orbits are forbidden.

Question 867. Use de-Broglie’s hypothesis to write the relation for the \( n^{th} \) radius of Bohr orbit interms of Bohr’s quantization condition of orbital angular momentum. 
Answer:
According to de-Broglie's hypothesis, the wavelength \( \lambda \) associated with the orbiting electron is: \[ \lambda = \frac{h}{p} = \frac{h}{mv_n} \] Only those orbits are stable where the electron's matter waves form stationary standing waves. For a circular orbit of radius \( r_n \), this condition is: \[ 2\pi r_n = n\lambda \quad \text{for } n = 1, 2, 3, \dots \] Substituting the expression for \( \lambda \): \[ 2\pi r_n = n\left(\frac{h}{mv_n}\right) \] Solving for the radius \( r_n \) of the \( n \)-th Bohr orbit: \[ r_n = \frac{nh}{2\pi m v_n} \] This relation expresses the orbit radius in terms of Planck's constant, the mass of the electron, and its velocity.
In simple words: The radius of the \( n \)-th orbit is given by the formula \( r_n = \frac{nh}{2\pi m v_n} \), which is derived by fitting a whole number of electron wavelengths around the orbit's path.

Exam Tip: Clearly show the substitution of de-Broglie's wavelength \( \lambda = \frac{h}{mv} \) into the standing wave formula \( 2\pi r = n\lambda \).

Question 868. (i) Define Ionization energy. What is its value for hydrogen atom ? 
(ii) How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge ?
Answer:
(i) **Ionization Energy:**
Ionization energy is defined as the minimum energy required to completely liberate an electron from its ground state in an atom to an infinite distance where it is no longer bound.
For a hydrogen atom, the ionization energy is: \[ E_{\text{ion}} = E_{\infty} - E_1 = 0 - (-13.6\text{ eV}) = +13.6\text{ eV} \]
(ii) **Effect of a heavier particle:**
The energy of the electron in the ground state of a hydrogen-like atom is given by: \[ E_1 = -\frac{m e^4}{8\varepsilon_0^2 h^2} \] This formula shows that the energy is directly proportional to the mass \( m \) of the orbiting particle:
\( \implies E_{\text{ion}} \propto m \delta \)
If the electron is replaced by a particle that has the same charge but is 200 times heavier, the ground state energy (and thus the energy required to remove it) will increase proportionally. Therefore, the ionization energy will become **200 times** its original value (which is \( 200 \times 13.6\text{ eV} = 2.72 \times 10^3\text{ eV} \)).
In simple words: (i) It is the energy needed to pull the electron completely away from the atom, which is \( 13.6\text{ eV} \) for hydrogen. (ii) Since a heavier particle is bound much more tightly, replacing the electron with a particle 200 times heavier increases the required ionization energy by 200 times.

Exam Tip: State the direct proportionality between the energy of the orbit and the mass of the orbiting particle to justify your answer for the second part.

Question 869. Draw a schematic arrangement of the Geiger – Marsden experiment for studying \( \alpha \)-particle scattering by a thin foil of gold. Describe briefly, by drawing trajectories of the scattered \( \alpha \)-particles, how this study can be used to estimate the size of the nucleus ? Draw a plot showing the number of particles scattered versus scattering angle \theta.
Answer:
**Schematic Setup:**
In this experiment, a highly collimated beam of energetic alpha particles from a radioactive source (like Bismuth-214) is directed at a very thin gold foil. The scattered particles are detected using a rotatable detector containing a zinc sulfide (ZnS) screen and a microscope.
**Estimation of Nuclear Size:**
By studying the trajectories of alpha particles, we observe that while most pass through undeflected, a very small fraction (about 1 in 8000) are scattered by more than \( 90^\circ \), and some bounce straight back (\( 180^\circ \)).
This allows us to calculate the distance of closest approach \( r_0 \) for a head-on collision, where the initial kinetic energy is completely converted into electrical potential energy: \[ \frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{r_0} \] This calculation yields a nuclear size limit of approximately \( 10^{-14}\text{ m} \).

**Plot of Number of Scattered Particles vs. Scattering Angle:**
The distribution shows that the number of scattered particles \( N(\theta) \) drops off sharply as the scattering angle increases:
In simple words: The experiment shoots alpha particles at a thin gold sheet. Since almost all go straight through but a tiny few bounce straight back, it proves the atom is mostly empty space with a tiny, dense, positive nucleus at the center.

Exam Tip: Be sure to draw both requested diagrams—the experimental setup and the trajectories—to ensure you get full credit for the drawing portion of the prompt.

Question 870. In Geiger- Marsden experiment, why is the most of the \alpha -Particles go straight through the foil and only a small fraction gets scattered at large angles ?
Answer:
For the vast majority of the incident alpha particles, the impact parameter \( b \) (sideways distance from the nucleus) is very large compared to the size of the nucleus. Because of this large distance, they experience a negligible electrostatic repulsive force from the positive nuclear charge and pass straight through the gold foil. Only those very few particles that pass close to the nucleus experience a strong enough repulsive force to be scattered at large angles.
In simple words: Since the nucleus is incredibly tiny compared to the size of the whole atom, almost all the alpha particles fly through empty space without getting close enough to be repelled.

Exam Tip: Use the term "impact parameter" to explain how the distance from the nucleus determines the scattering angle of the particles.

Question 871. In Geiger-Marsden experiment, draw the trajectories traced by \alpha -Particles in the Coulomb’s field of target. 
Answer:
The trajectories of alpha particles scattered at different distances (impact parameters) from the target nucleus are shown in the diagram below:
This trajectory distribution demonstrates that the deflection angle increases as the impact parameter decreases.
In simple words: The diagram shows how particles passing far away travel in straight lines, while those passing closer are bent into curves, and head-on particles bounce straight back.

Exam Tip: Label the target nucleus and draw the different curves clearly to show how the scattering angle depends on the starting height.

Question 872. Using Bohr’s postulates, derive the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence, derive the expression for the orbital velocity and orbital period of the electron moving in the \( n^{th} \) orbit of hydrogen atom. CBSE (F)-2017,2014,2012,2011,(AI)-2015,2014,2013,(D)-2013
Answer:
Let \( m \), \( e \), \( v \), and \( r \) represent the mass, charge, orbital velocity, and orbit radius of the electron in a hydrogen-like atom of atomic number \( Z \).

**1. Radius of stationary orbits:**
The electrostatic force of attraction between the electron and the nucleus provides the necessary centripetal force: \[ \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r^2} \]
\( \implies mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r} \) ----(1)
According to Bohr's quantization condition: \[ mvr = n\frac{h}{2\pi} \implies v = \frac{nh}{2\pi m r} \] Substituting \( v \) into equation (1): \[ m \left(\frac{nh}{2\pi m r}\right)^2 = \frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r} \] \[ \frac{n^2 h^2}{4\pi^2 m r^2} = \frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r} \]
\( \implies r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2} \) ----(2)

**2. Total Energy of Electron (\( E \)):**
The total energy is the sum of kinetic energy (\( E_K \)) and electrostatic potential energy (\( U \)): \[ E_K = \frac{1}{2}mv^2 = \frac{1}{8\pi\varepsilon_0} \frac{Ze^2}{r} \] \[ U = -\frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r} \] \[ E = E_K + U = \frac{1}{8\pi\varepsilon_0} \frac{Ze^2}{r} - \frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r} = -\frac{1}{8\pi\varepsilon_0} \frac{Ze^2}{r} \]
Substituting the expression for \( r_n \) from equation (2): \[ E_n = -\frac{1}{8\pi\varepsilon_0} Ze^2 \left( \frac{\pi m Z e^2}{\varepsilon_0 n^2 h^2} \right) = -\frac{m Z^2 e^4}{8\varepsilon_0^2 n^2 h^2} \]
For hydrogen (\( Z = 1 \)): \[ E_n = -\frac{m e^4}{8\varepsilon_0^2 n^2 h^2} = -\frac{13.6}{n^2}\text{ eV} \]

**3. Orbital Velocity (\( v \)):**
Dividing the centripetal equation (1) by the quantization condition: \[ \frac{mv^2}{mvr} = \left(\frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r^2}\right) \div \left(\frac{nh}{2\pi}\right) \]
\( \implies v = \frac{Z e^2}{2\varepsilon_0 n h} \)
For hydrogen (\( Z = 1 \)): \[ v_n = \frac{e^2}{2\varepsilon_0 n h} = \left(\frac{c}{137}\right)\frac{1}{n} \]

**4. Orbital Period (\( T \)):**
The time period \( T \) for one complete revolution is: \[ T = \frac{2\pi r}{v} \]
Substituting the expressions for \( r_n \) and \( v_n \): \[ T = \frac{2\pi \left(\frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}\right)}{\frac{Z e^2}{2\varepsilon_0 n h}} = \frac{4\varepsilon_0^2 n^3 h^3}{m Z^2 e^4} \]
Thus, the orbital period is proportional to \( n^3 \).
In simple words: By balancing centripetal force with electrostatic attraction and applying angular momentum quantization, we calculate that the orbit radius scales as \( n^2 \), the velocity decreases as \( 1/n \), the total energy is negative and decreases as \( 1/n^2 \), and the orbital time period grows as \( n^3 \).

Exam Tip: This is a long and highly graded derivation. Memorize key intermediate steps, particularly the expressions for \( r \) and \( v \), to ensure you can reconstruct them under exam pressure.

Question 873. (a) Explain the origin of spectral series/ lines of hydrogen atom using Bohr’s atomic model.
(b) Draw the energy level diagram showing how the line spectra corresponding to Lyman/Balmer series occur due to transition between energy levels in a hydrogen atom. 
Answer:
(a) **Origin of Spectral Lines:**
According to Bohr's model, an electron in a hydrogen atom can exist only in specific quantized energy levels. When an electron transitions from a higher energy level \( n_2 \) (energy \( E_2 \)) to a lower energy level \( n_1 \) (energy \( E_1 \)), it emits a photon. The frequency \( \nu \) of the emitted photon is: \[ h\nu = E_2 - E_1 \] The wave number \( \bar{\nu} = \frac{1}{\lambda} \) is: \[ \bar{\nu} = \frac{E_2 - E_1}{hc} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7\text{ m}^{-1} \)). Different series are obtained depending on the value of the final level \( n_1 \):
- **Lyman Series (\( n_1 = 1 \)):** Occurs in the ultraviolet region (not visible) for \( n_2 = 2, 3, 4, \dots \)
- **Balmer Series (\( n_1 = 2 \)):** Occurs in the visible region for \( n_2 = 3, 4, 5, \dots \)
- **Paschen Series (\( n_1 = 3 \)):** Occurs in the near-infrared region for \( n_2 = 4, 5, 6, \dots \)
- **Brackett Series (\( n_1 = 4 \)):** Occurs in the infrared region for \( n_2 = 5, 6, 7, \dots \)
- **Pfund Series (\( n_1 = 5 \)):** Occurs in the far-infrared region for \( n_2 = 6, 7, 8, \dots \)

(b) **Energy Level Diagram:**
The transitions for the Lyman and Balmer series are shown in the energy level diagram below:
In simple words: (a) When an electron falls from a outer level to an inner level, it releases light of a specific color (wavelength). Grouping these lines based on where they land creates the spectral series. (b) The diagram shows these jumps, with Lyman landing on \( n=1 \) and Balmer landing on \( n=2 \).

Exam Tip: Be sure to write the correct energy values (\( -13.6\text{ eV} \), \( -3.4\text{ eV} \), etc.) next to each line in the energy level diagram to show accuracy.

Question 874. Draw a plot of potential energy of a pair of nucleons as a function of their separations.
(i) Write two important conclusions that can be drawn from the graph.
(ii) What is the significance of negative potential energy in the graph drawn ? 
Answer:
The potential energy curve of a pair of nucleons as a function of their separation distance \( r \) is shown below:
(i) **Two Important Conclusions:**
1. **Repulsive behavior at very short range:** For separations \( r < r_0 \) (\( \approx 0.8\text{ fm} \)), the potential energy rises very steeply. This indicates a strong repulsive force between the nucleons, preventing them from collapsing into each other.
2. **Attractive behavior at medium range:** For separations \( r > r_0 \), the potential energy is negative, representing an attractive force that holds them together. This attraction falls rapidly to zero for distances greater than a few Fermi.

(ii) **Significance of Negative Potential Energy:**
The negative potential energy indicates that the state is stable and bound. It shows that energy must be supplied from outside the system to break the nucleons apart, which represents the source of nuclear binding energy.
In simple words: The graph shows that if nucleons get too close (less than 0.8 fm), they strongly repel each other. If they are slightly further apart, they attract, creating a stable, bound nucleus where the negative energy holds them together.

Exam Tip: Clearly label the equilibrium separation point \( r_0 \approx 0.8\text{ fm} \), which corresponds to the minimum potential energy of the system.

Question 875. Draw a plot of binding energy per nucleon (B.E/A) as a function of mass number A.
(a) Write salient features of this curve.
(b) Write two important conclusions that can be drawn regarding the nature of nuclear force.
(c) Use this graph to explain the release of energy in both the processes of nuclear fission and fusion.
Answer:
The binding energy per nucleon (\( E_{bn} \)) as a function of mass number \( A \) is shown in the curve below:
(a) **Salient Features:**
1. **Constant stability for medium nuclei:** For mass numbers between \( 30 < A < 170 \), the binding energy per nucleon is practically constant, maintaining a high value of approximately \( 8.0\text{ MeV} \).
2. **Maximum peak:** The curve reaches its absolute maximum of about \( 8.75\text{ MeV/nucleon} \) at \( A = 56 \), which corresponds to the exceptionally stable \( ^{56}Fe \) nucleus.
3. **Lower values at ends:** The binding energy per nucleon is relatively low for both very light nuclei (\( A < 30 \)) and very heavy nuclei (\( A > 170 \)), making them less stable.

(b) **Conclusions on the Nature of Nuclear Force:**
1. **Short-range saturation:** The constancy of \( E_{bn} \) over the range \( 30 < A < 170 \) indicates that a nucleon interacts only with its immediate neighbors. This demonstrates the saturation property of short-range nuclear forces.
2. **Strong attraction:** The high value of several MeV per nucleon indicates that the nuclear force is exceptionally strong, far exceeding electrostatic or gravitational forces.

(c) **Explanation of Energy Release:**
1. **Nuclear Fission:** When a heavy, unstable nucleus (e.g., \( A \approx 240 \) with \( E_{bn} \approx 7.6\text{ MeV} \)) splits into two medium-sized fragments (e.g., \( A \approx 120 \) with \( E_{bn} \approx 8.5\text{ MeV} \)), the binding energy per nucleon increases. The product nucleons are more tightly bound, resulting in a mass loss that is released as energy.
2. **Nuclear Fusion:** When two very light nuclei (e.g., \( A \le 10 \) with low \( E_{bn} \)) combine to form a single heavier, more stable nucleus, the binding energy per nucleon increases sharply. This large gain in binding energy releases a massive amount of energy.
In simple words: The graph shows that medium-sized elements like Iron-56 are the most stable because their particles are held together with the most energy. Very light elements can fuse together, and very heavy elements can split apart (fission), both moving toward the stable middle and releasing energy in the process.

Exam Tip: Be prepared to explain how the flatness of the curve in the middle range proves that the nuclear force is "short-ranged" (saturation property).

Question 876. What characteristic property of nuclear force explains the consistency of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30< A< 170 ?
Answer:
The consistency of binding energy per nucleon is explained by the **saturation property** of nuclear forces, which is a direct consequence of their short-range nature. A nucleon inside a medium-sized nucleus interacts only with its immediate neighboring nucleons. Adding more nucleons to a nucleus beyond its immediate neighborhood does not increase the binding energy of the existing nucleons, keeping the average binding energy per nucleon constant.
In simple words: Because the nuclear force only works over very short distances, a single particle can only feel the pull of its closest neighbors. Adding more particles to the outside of a medium-sized nucleus doesn't affect the inside particles, keeping the average binding strength consistent.

Exam Tip: Use the term "saturation of nuclear forces" and explain that it arises because of their short-range character to secure full marks.

Question 877. Give the reason for the decrease of binding energy per nucleon for nuclei with high numbers. 
Answer:
As the mass number of a nucleus becomes very large, the number of protons increases significantly. Since the attractive nuclear force is short-ranged and saturates, it does not grow with the size of the nucleus. However, the repulsive electrostatic Coulomb force between the protons is long-ranged and acts across the entire nucleus. This growing repulsive force eventually begins to destabilize the heavy nucleus, causing the average binding energy per nucleon to decrease.
In simple words: In very large nuclei, the positive charges of the many protons repel each other across long distances, while the holding force is only short-range. This push-apart force weakens the average binding strength of the heavy nucleus.

Exam Tip: Contrast the "short-range attractive nuclear force" with the "long-range repulsive Coulomb force" to provide a complete scientific explanation.

Question 878. The figure shows the plot of binding energy per nucleon as a function of mass number A. Point out, giving reasons, the two processes (in terms of A,B,C,D and E) , one of which can occur due to nuclear fission and the other due to nuclear fusion. 
Answer:
Let us identify the processes based on the general binding energy curve:
1. **Nuclear Fission of E into C and D:**
The heavy nucleus \( E \) has a lower binding energy per nucleon. When it splits into two lighter fragments \( C \) and \( D \), the binding energy per nucleon increases, making the product nuclei more stable and releasing energy.

2. **Nuclear Fusion of A and B into D (or C):**
The very light nuclei \( A \) and \( B \) have low binding energies per nucleon. When they fuse together to form a heavier, more stable nucleus like \( D \), there is a sharp increase in the binding energy per nucleon, which releases a massive amount of energy.
In simple words: Fission occurs when heavy nucleus E splits into the more stable, medium-sized C and D. Fusion occurs when the light nuclei A and B join to form the more stable, heavier nucleus D.

Exam Tip: State that both processes occur because the system always tends to transition toward states with higher binding energy per nucleon to achieve stability.

Question 879. State the law of radioactive decay.
(i) Derive the mathematical expression for law of radioactive decay for a sample of radioactive nucleus.
(ii) Plot a graph showing the number (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half-life T.
Answer:
**Law of Radioactive Decay:**
The rate of decay (or disintegration) of a radioactive sample at any instant is directly proportional to the total number of active, undecayed nuclei present in the sample at that moment.

(i) **Mathematical Derivation:**
Let \( N(t) \) be the number of undecayed nuclei present in a sample at time \( t \). If \( dN \) nuclei disintegrate in a small time interval \( dt \), then the rate of decay is \( -\frac{dN}{dt} \). According to the law: \[ -\frac{dN}{dt} \propto N \]
\( \implies \frac{dN}{dt} = -\lambda N \) ----(1)
where \( \lambda \) is the decay constant of the radioactive substance. Rearranging the terms of equation (1): \[ \frac{dN}{N} = -\lambda dt \]
Integrating both sides of the equation: \[ \int \frac{dN}{N} = -\lambda \int dt \] \[ \ln N = -\lambda t + C \]
At \( t = 0 \), let the initial number of undecayed nuclei be \( N_0 \). This gives the integration constant: \[ \ln N_0 = C \] Substituting \( C \) back into the integrated equation: \[ \ln N - \ln N_0 = -\lambda t \] \[ \ln\left(\frac{N}{N_0}\right) = -\lambda t \] Taking the exponential of both sides: \[ \frac{N}{N_0} = e^{-\lambda t} \]
\( \implies N = N_0 e^{-\lambda t} \)
This is the exponential decay law.

(ii) **Decay Curve:**
The exponential decrease of undecayed nuclei over time is plotted below:
In simple words: The law says that the more radioactive material you have, the faster it decays. Over time, the amount of remaining material drops exponentially, cutting in half with each half-life period.

Exam Tip: Be sure to include the negative sign in the starting differential equation \( \frac{dN}{dt} = -\lambda N \) and state that it signifies the decrease in the number of undecayed nuclei over time.

Question 880. Define the terms half-life period & decay constant of a radioactive substance. Write their S.I. units. Establish the relation between them.
Answer:
**Half-Life Period (\( T \)):**
The half-life of a radioactive substance is defined as the time taken for the number of active, undecayed nuclei in a sample to reduce to exactly half of its initial value. Its S.I. unit is the **second (s)**.

**Decay Constant (\( \lambda \)):**
The decay constant is defined as the reciprocal of the time interval in which the number of active nuclei left undecayed reduces to \( 1/e \) times (about 36.8%) of its original value. Its S.I. unit is **\( \text{second}^{-1} \) (\( \text{s}^{-1} \))**.

**Derivation of Relationship:**
According to the exponential decay law: \[ N = N_0 e^{-\lambda t} \]
By definition, when the elapsed time is equal to one half-life (\( t = T \)), the remaining number of nuclei is: \[ N = \frac{N_0}{2} \]
Substituting these conditions into the decay law: \[ \frac{N_0}{2} = N_0 e^{-\lambda T} \] \[ \frac{1}{2} = e^{-\lambda T} \] Taking the natural logarithm of both sides: \[ \ln(2) = \lambda T \] \[ \lambda T = 0.6931 \]
\( \implies T = \frac{0.6931}{\lambda} \)
This is the relation between the half-life and the decay constant.
In simple words: Half-life is the time it takes for half of your radioactive sample to decay. The decay constant measures how quickly it disintegrates. They are related by the constant value \( 0.6931 \) divided by the decay rate.

Exam Tip: Memorize the final formula \( T = \frac{0.693}{\lambda} \), as it is highly useful for numerical calculations throughout the physics curriculum.

Question 881. Define the term mean life of a radioactive nuclide. How is the mean life of a given radioactive nucleus related to the decay constant and Half-life ? 
Answer:
**Mean Life (\( \tau \)):**
The mean life (or average life) of a radioactive substance is defined as the sum of the lifetimes of all the individual nuclei in a sample divided by the total number of nuclei initially present.
Mathematically, it is equal to the reciprocal of the decay constant \( \lambda \): \[ \tau = \frac{1}{\lambda} \]
**Relationship with Half-Life (\( T \)):**
We know the relationship between half-life and decay constant is: \[ T = \frac{\ln 2}{\lambda} = 0.693 \tau \] Therefore, the mean life \( \tau \) can be related to the half-life \( T \) by: \[ \tau = \frac{T}{0.693} \approx 1.44 T \]
Thus, the mean life is approximately \( 1.44 \) times the half-life of the radioactive substance.
In simple words: Mean life is the average lifespan of a radioactive nucleus. It is equal to \( 1.44 \) times the half-life of the substance.

Exam Tip: Clearly distinguish between mean life (\( \tau = 1.44 T \)) and half-life (\( T \)) to avoid confusing these two concepts in descriptive answers.

Question 882. Define activity of a radioactive substance and write its S.I. unit. Plot a graph showing variation of activity of a given radioactive sample with time(F)
Answer:
**Activity (\( R \)):**
The activity of a radioactive substance is defined as the rate of disintegration (or decay) of the nuclei in the sample at any given instant. Mathematically, it is expressed as: \[ R = -\frac{dN}{dt} = \lambda N \] The S.I. unit of activity is the **Becquerel (Bq)**, where \( 1\text{ Bq} = 1\text{ disintegration per second} \). Another common non-SI unit is the Curie (Ci).

**Activity vs. Time Graph:**
Since \( R \propto N \) and \( N \) decreases exponentially with time, the activity \( R \) also decreases exponentially according to \( R = R_0 e^{-\lambda t} \).
In simple words: Activity is the number of atomic decays happening every second in a sample. Like the amount of material itself, the activity drops exponentially over time.

Exam Tip: Name the S.I. unit explicitly as the "Becquerel" and provide its definition to secure complete credit for unit-related questions.

Question 883. Show that the decay rate ‘R’ of a sample of a radioactive nuclide is related to the number of radioactive nuclei ‘N’ at the same instant by the expression \( R = \lambda N \) & \( \frac{dR}{dt} \propto \frac{1}{T^2} \) 
Answer:
According to the radioactive decay law, the number of undecayed nuclei is: \[ N = N_0 e^{-\lambda t} \]
The decay rate \( R \) (or activity) is defined as: \[ R = -\frac{dN}{dt} = -\frac{d}{dt}(N_0 e^{-\lambda t}) = -N_0 (-\lambda e^{-\lambda t}) = \lambda (N_0 e^{-\lambda t}) \]
\( \implies R = \lambda N \) ----(1)
This proves the first part of the expression.

To find the rate of change of the decay rate: \[ \frac{dR}{dt} = \frac{d}{dt}(\lambda N) = \lambda \frac{dN}{dt} \] Substituting \( \frac{dN}{dt} = -\lambda N \): \[ \frac{dR}{dt} = -\lambda^2 N \] Taking the magnitude of this rate: \[ \left|\frac{dR}{dt}\right| = \lambda^2 N \]
We know that the decay constant \( \lambda \) is related to the half-life \( T \) by \( \lambda = \frac{\ln 2}{T} \). Substituting this: \[ \left|\frac{dR}{dt}\right| = \left(\frac{\ln 2}{T}\right)^2 N = \frac{(\ln 2)^2 N}{T^2} \] Since \( (\ln 2)^2 \) and \( N \) are constant at any given instant:
\( \implies \left|\frac{dR}{dt}\right| \propto \frac{1}{T^2} \)
This proves the second part of the relation.
In simple words: The rate of decay is equal to the decay constant times the number of remaining nuclei. By taking the derivative, we show that how fast the decay rate changes is inversely proportional to the square of the half-life.

Exam Tip: Be sure to write the full steps for the derivative of the exponential function \( e^{-\lambda t} \) to avoid any deduction of marks.

Question 884. A radioactive sample having \( N \) nuclei has activity R. Write down an expression for its half-life in terms of R and N 
Answer:
The relationship between activity \( R \), decay constant \( \lambda \), and the number of nuclei \( N \) is: \[ R = \lambda N \implies \lambda = \frac{R}{N} \]
The half-life \( T \) of the sample is related to the decay constant by: \[ T = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \] Substituting the value of \( \lambda \) from above: \[ T = \frac{0.693 \cdot N}{R} \]
This is the required expression for half-life in terms of \( R \) and \( N \).
In simple words: The half-life is equal to \( 0.693 \) times the number of nuclei divided by the current activity of the sample.

Exam Tip: Be sure to write the final formula clearly, defining both \( R \) and \( N \) to make your answer complete.

Question 885. What is nuclear fission and fusion ? Give one representative equation of each.
Answer:
**Nuclear Fission:**
Nuclear fission is the process in which a heavy, unstable nucleus splits into two or more lighter, more stable nuclei when bombarded with slow neutrons, accompanied by the release of a massive amount of energy and several neutrons.
*Representative Equation:* \[ _{92}^{235}U + _0^1n \to _{56}^{144}Ba + _{36}^{89}Kr + 3_0^1n + Q \]
**Nuclear Fusion:**
Nuclear fusion is the process in which two or more very light, high-energy nuclei combine at extremely high temperatures and pressures to form a single heavier, more stable nucleus, releasing a tremendous amount of energy.
*Representative Equation:* \[ _1^2H + _1^2H \to _2^3He + _0^1n + 3.27\text{ MeV} \]
In simple words: Fission is the splitting of a heavy atom (like Uranium) into smaller ones. Fusion is the joining of light atoms (like Hydrogen isotopes) to make a heavier one. Both processes release huge amounts of energy.

Exam Tip: Be sure to include the energy term (\( Q \) or the specific MeV value) in both equations, as energy release is a defining feature of these reactions.

Question 886. What is nuclear reactor ? Draw a labelled diagram of a nuclear reactor. Write its principle and explain its working.
Answer:
**Nuclear Reactor:**
A nuclear reactor is a device designed to initiate, sustain, and control a self-sustaining nuclear fission chain reaction to generate heat, which is then used to produce electricity.

**Principle:**
It operates on the principle of a **controlled nuclear fission chain reaction**. By using control rods to absorb excess neutrons, the rate of fission is kept constant, ensuring a steady release of energy.

**Labelled Schematic Diagram:**
The core components of a nuclear power plant are represented below:
**Working:**
1. **Fuel:** Enriched uranium (\( ^{235}U \)) is placed in the reactor core as fuel to undergo fission.
2. **Moderator:** Graphite or heavy water (\( D_2O \)) is used to slow down the fast neutrons produced in fission, converting them into thermal neutrons to maintain the reaction.
3. **Control Rods:** Cadmium or Boron rods, which absorb neutrons efficiently, are inserted or withdrawn from the core to control the rate of fission.
4. **Coolant:** Water or liquid sodium is circulated through the core to absorb the heat produced by fission, which is then used to generate steam to rotate turbines for electricity.
In simple words: A nuclear reactor is a machine that splits uranium atoms in a controlled way to create heat. This heat boils water to make steam, which spins a turbine to generate clean electrical energy.

Exam Tip: Be sure to explicitly list the roles of the three main components—Moderator (slows neutrons), Control Rods (absorbs neutrons), and Coolant (transfers heat)—as they are highly graded.

Question 887. Find the relation between the three wavelengths \( \lambda_1 \), \( \lambda_2 \) and \( \lambda_3 \) from the energy level diagram shown below. 
Answer:
Let \( E_A \), \( E_B \), and \( E_C \) be the energies of the levels \( A \), \( B \), and \( C \) respectively, as shown in the diagram. The transitions are: 1. From level \( C \) to \( B \), emitting a photon of wavelength \( \lambda_1 \): \[ E_C - E_B = \frac{hc}{\lambda_1} \] ----(1)
2. From level \( B \) to \( A \), emitting a photon of wavelength \( \lambda_2 \): \[ E_B - E_A = \frac{hc}{\lambda_2} \] ----(2)
3. From level \( C \) to \( A \), emitting a photon of wavelength \( \lambda_3 \): \[ E_C - E_A = \frac{hc}{\lambda_3} \] ----(3)
Adding equation (1) and equation (2): \[ (E_C - E_B) + (E_B - E_A) = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2} \] \[ E_C - E_A = hc \left( \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \right) \] ----(4)
Equating the expressions for \( E_C - E_A \) from equation (3) and equation (4): \[ \frac{hc}{\lambda_3} = hc \left( \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \right) \]
\( \implies \frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \)

\( \implies \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} \)
This is the required relation between the three wavelengths.
In simple words: The energy of the long jump (C to A) is the sum of the energies of the two smaller jumps (C to B and B to A). Since energy is inversely proportional to wavelength, the reciprocal of the long wavelength is the sum of the reciprocals of the two shorter ones.

Exam Tip: Clearly show the addition of equations (1) and (2) to demonstrate how \( E_B \) cancels out, proving the final reciprocal relationship.

Question 888. The figure shows energy level diagram of hydrogen atom. 
(i) Find out the transition which results in the emission of a photon of wavelength 496 nm
(ii) Which transition corresponds to the emission of radiation of maximum wavelength ? Justify your answer.
Answer:
The energy level diagram is shown below:
(i) **Transition for \( 496\text{ nm} \):**
The energy of the emitted photon with \( \lambda = 496\text{ nm} \) is: \[ \Delta E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{496 \times 10^{-9}}\text{ J} = 4.0 \times 10^{-19}\text{ J} \] Converting this energy value into electron-volts: \[ \Delta E = \frac{4.0 \times 10^{-19}}{1.6 \times 10^{-19}}\text{ eV} \approx 2.5\text{ eV} \]
For a hydrogen atom, the level energies are \( E_1 = -13.6\text{ eV} \), \( E_2 = -3.4\text{ eV} \), \( E_3 = -1.51\text{ eV} \), and \( E_4 = -0.85\text{ eV} \). The transition that corresponds to an energy change of \( 2.5\text{ eV} \) is: \[ E_4 - E_2 = -0.85\text{ eV} - (-3.4\text{ eV}) = 2.55\text{ eV} \] Thus, the transition is from **\( n = 4 \) to \( n = 2 \)** (labeled as **B**).

(ii) **Transition for Maximum Wavelength:**
Since \( \lambda \propto \frac{1}{\Delta E} \), the maximum wavelength corresponds to the minimum energy difference. Comparing the transitions shown: - Transition A (\( 4 \to 1 \)): \( \Delta E = 12.75\text{ eV} \) - Transition B (\( 4 \to 2 \)): \( \Delta E = 2.55\text{ eV} \) - Transition C (\( 3 \to 2 \)): \( \Delta E = -1.51 - (-3.4) = 1.89\text{ eV} \)
Transition C (\( 3 \to 2 \)) has the minimum energy gap. Therefore, **Transition C** corresponds to the emission of radiation of maximum wavelength.
In simple words: (i) A \( 496\text{ nm} \) photon has \( 2.5\text{ eV} \) of energy, which matches the jump from level 4 to level 2 (transition B). (ii) The longest wavelength comes from the smallest energy jump, which is the transition from level 3 to level 2 (transition C).

Exam Tip: Explicitly mention that energy and wavelength are inversely related to justify your selection for the maximum wavelength transition.

Question 889. A hydrogen atom initially in its ground state absorbs a photon and is in the excited state with energy 12.5 eV. Calculate the longest wavelength of the radiation emitted and identify the series to which it belongs . (Rydberg constant, \( R = 1.1 \times 10^7\text{ m}^{-1} \)) 
Answer:
The energy of the electron in the ground state of hydrogen is \( -13.6\text{ eV} \). After absorbing a photon of \( 12.5\text{ eV} \), the energy of the excited state is: \[ E_n = -13.6\text{ eV} + 12.5\text{ eV} = -1.1\text{ eV} \]
Using the energy formula \( E_n = -\frac{13.6}{n^2}\text{ eV} \): \[ -1.1 = -\frac{13.6}{n^2} \implies n^2 = \frac{13.6}{1.1} \approx 12.36 \]
Since \( n \) must be an integer, the closest integer is \( n = 3 \) (with energy \( -1.51\text{ eV} \)). Let us assume the electron is excited to \( n = 3 \).
The longest wavelength emitted from \( n = 3 \) corresponds to the smallest energy jump, which is the transition from \( n = 3 \) to \( n = 2 \). This belongs to the Balmer series. Using the Rydberg formula: \[ \frac{1}{\lambda_{\max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5R}{36} \]
\( \implies \lambda_{\max} = \frac{36}{5R} \)
Substituting \( R = 1.1 \times 10^7\text{ m}^{-1} \): \[ \lambda_{\max} = \frac{36}{5 \times 1.1 \times 10^7} = \frac{36}{5.5 \times 10^7} \approx 6.545 \times 10^{-7}\text{ m} = 6545\text{ \AA} \]
This transition belongs to the **Balmer series** (red light).
In simple words: The absorbed energy excites the electron to level 3. The longest wave it can emit when falling back is the small jump from 3 to 2, which has a wavelength of \( 6545\text{ \AA} \) and belongs to the Balmer series.

Exam Tip: Be sure to state both the calculated numerical value of the wavelength and the name of the series (Balmer) as requested.

Question 890. Using Rydberg’s formula, calculate the longest wavelengths belonging to Lyman and Balmer series. In which region f hydrogen spectrum do these transmission lie ? (Given, \( R = 1.1 \times 10^7\text{ m}^{-1} \)) 
Answer:
The Rydberg formula is: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
**1. Lyman Series (Longest Wavelength):**
The longest wavelength corresponds to the transition from the nearest higher orbit, \( n_2 = 2 \) to \( n_1 = 1 \): \[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \]
\( \implies \lambda_L = \frac{4}{3R} \)
Substituting \( R = 1.1 \times 10^7\text{ m}^{-1} \): \[ \lambda_L = \frac{4}{3 \times 1.1 \times 10^7} \approx 1.21 \times 10^{-7}\text{ m} = 1210\text{ \AA} \]
This transition lies in the **Ultraviolet** region.

**2. Balmer Series (Longest Wavelength):**
The longest wavelength corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \): \[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36} \]
\( \implies \lambda_B = \frac{36}{5R} \)
Substituting \( R = 1.1 \times 10^7\text{ m}^{-1} \): \[ \lambda_B = \frac{36}{5 \times 1.1 \times 10^7} \approx 6.545 \times 10^{-7}\text{ m} = 6545\text{ \AA} \]
This transition lies in the **Visible** region.
In simple words: The longest wave in the Lyman series is \( 1210\text{ \AA} \) (ultraviolet), while the longest in the Balmer series is \( 6545\text{ \AA} \) (visible).

Exam Tip: Clearly state the spectral regions—Ultraviolet for Lyman and Visible for Balmer—as they are required for complete marks.

Question 891. The ground state energy of hydrogen atom is -13.6 eV. 
(i) what is the kinetic energy of an electron in the 2nd excited state ?
(ii) If the electron jumps to the ground state from 2nd excited state, calculate the wavelength of the spectral line emitted.
Answer:
(i) The second excited state corresponds to the third energy level, \( n = 3 \). The total energy in this state is: \[ E_3 = -\frac{13.6}{3^2} = -1.51\text{ eV} \]
Since kinetic energy \( E_K \) is equal to the negative of total energy: \[ E_K = -E_3 = -(-1.51\text{ eV}) = +1.51\text{ eV} \]
(ii) The transition is from \( n_2 = 3 \) to \( n_1 = 1 \). The energy change \( \Delta E \) is: \[ \Delta E = E_3 - E_1 = -1.51\text{ eV} - (-13.6\text{ eV}) = 12.09\text{ eV} \] Converting this energy into Joules: \[ \Delta E = 12.09 \times 1.6 \times 10^{-19}\text{ J} = 1.9344 \times 10^{-18}\text{ J} \]
The wavelength \( \lambda \) of the emitted photon is: \[ \lambda = \frac{hc}{\Delta E} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9344 \times 10^{-18}}\text{ m} \] \[ \lambda = \frac{1.98 \times 10^{-25}}{1.9344 \times 10^{-18}} \approx 1.02 \times 10^{-7}\text{ m} = 1020\text{ \AA} \]
Thus, the wavelength of the spectral line is **\( 1020\text{ \AA} \)**.
In simple words: (i) In the 3rd level, the electron's kinetic energy is \( +1.51\text{ eV} \). (ii) When it falls from level 3 to level 1, it emits ultraviolet light with a wavelength of \( 1020\text{ \AA} \).

Exam Tip: Ensure you use \( n = 3 \) for the "second excited state" to prevent introducing errors at the start of your calculations.

Question 892. Two different radioactive elements with half lives \( T_1 \) and \( T_2 \) have \( N_1 \) and \( N_2 \) undecayed atoms respectively present at a given instant. Derive an expression for the ratio of their activities at this instant in terms of \( N_1 \) & \( N_2 \) 
Answer:
The activity \( R \) of a radioactive sample is given by the relation: \[ R = \lambda N \] where \( \lambda \) is the decay constant and \( N \) is the number of active nuclei.
For the first element: \[ R_1 = \lambda_1 N_1 \] For the second element: \[ R_2 = \lambda_2 N_2 \]
Taking the ratio of their activities: \[ \frac{R_1}{R_2} = \frac{\lambda_1 N_1}{\lambda_2 N_2} \] ----(1)
We know that the decay constant \( \lambda \) is related to the half-life \( T \) by: \[ \lambda = \frac{\ln 2}{T} \] Substituting the expressions for \( \lambda_1 \) and \( \lambda_2 \) into equation (1): \[ \frac{R_1}{R_2} = \frac{\left(\frac{\ln 2}{T_1}\right) N_1}{\left(\frac{\ln 2}{T_2}\right) N_2} \]
\( \implies \frac{R_1}{R_2} = \frac{N_1}{N_2} \times \frac{T_2}{T_1} \)
This is the required expression for the ratio of their activities.
In simple words: The ratio of the activities of two samples is equal to the ratio of their particle counts multiplied by the inverse ratio of their half-lives.

Exam Tip: Write down the intermediate step showing the substitution of \( \lambda = \frac{0.693}{T} \) to make the derivation logical and easy to follow.

Question 893. Half life of \( _{92}^{238}U \) against \alpha -decay is \( 4.5 \times 10^9 \) years. Calculate the activity of 1 g sample of \( _{92}^{238}U \). (Given Avogadro’s number = \( 6 \times 10^{26} \) atoms/ Kmol ) 
Answer:
**1. Convert Half-Life to Seconds:**
\[ T = 4.5 \times 10^9\text{ years} = 4.5 \times 10^9 \times 365 \times 24 \times 3600\text{ s} \] \[ T \approx 1.42 \times 10^{17}\text{ s} \]
**2. Calculate the Number of Atoms (\( N \)) in 1 g of Uranium-238:**
Using Avogadro's number (\( 6 \times 10^{23}\text{ atoms/mol} \) or \( 6 \times 10^{26}\text{ atoms/kmol} \)): \[ N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{1}{238} \times 6 \times 10^{23}\text{ atoms} \] \[ N \approx 2.52 \times 10^{21}\text{ atoms} \]
**3. Calculate the Activity (\( R \)):**
The decay constant is: \[ \lambda = \frac{0.693}{T} = \frac{0.693}{1.42 \times 10^{17}\text{ s}} \approx 4.88 \times 10^{-18}\text{ s}^{-1} \] The activity \( R \) is: \[ R = \lambda N = 4.88 \times 10^{-18} \times 2.52 \times 10^{21}\text{ Bq} \] \[ R \approx 1.23 \times 10^4\text{ Bq} \]
Thus, the activity of the 1 g sample is approximately **\( 1.23 \times 10^4\text{ Bq} \)**.
In simple words: One gram of Uranium-238 contains \( 2.52 \times 10^{21} \) atoms. Because it decays extremely slowly, only about \( 12,300 \) atoms disintegrate each second in the entire sample.

Exam Tip: Always pay attention to the unit of Avogadro's number provided (atoms/Kmol requires converting the mass to kg, or simply using \( 6 \times 10^{23}\text{ atoms/mol} \) for grams).

Question 894. A radioactive sample contains 2.2 mg of pure \( _6^{11}C \) which has half-life period of 1224 seconds. Calculate : (i) the number of atoms present initially. 
(ii) the activity when 5 \mu g of the sample will be left.
Answer:
We are given:
Initial mass \( m_0 = 2.2\text{ mg} = 2.2 \times 10^{-3}\text{ g} \)
Half-life \( T = 1224\text{ s} \)

(i) **Initial Number of Atoms (\( N_0 \)):**
Molar mass of Carbon-11 is \( 11\text{ g/mol} \). \[ N_0 = \frac{m_0}{M} \times N_A = \frac{2.2 \times 10^{-3}\text{ g}}{11\text{ g/mol}} \times 6 \times 10^{23}\text{ atoms/mol} \] \[ N_0 = 0.2 \times 10^{-3} \times 6 \times 10^{23} = 1.2 \times 10^{20}\text{ atoms} \]
(ii) **Activity \( R \) when \( 5\,\mu\text{g} \) is left:**
Remaining mass \( m = 5\,\mu\text{g} = 5 \times 10^{-6}\text{ g} \). First, find the number of remaining atoms \( N \): \[ N = \frac{m}{M} \times N_A = \frac{5 \times 10^{-6}\text{ g}}{11\text{ g/mol}} \times 6 \times 10^{23} \approx 2.73 \times 10^{17}\text{ atoms} \] Now, calculate the activity: \[ R = \lambda N = \frac{0.693}{T} \times N = \frac{0.693}{1224\text{ s}} \times 2.73 \times 10^{17} \] \[ R \approx 5.66 \times 10^{-4} \times 2.73 \times 10^{17}\text{ Bq} \approx 1.55 \times 10^{14}\text{ Bq} \]
Thus, the activity is **\( 1.55 \times 10^{14}\text{ Bq} \)**.
In simple words: (i) Initially, there are \( 1.2 \times 10^{20} \) carbon atoms. (ii) When only \( 5\text{ micrograms} \) of carbon remains, there are still enough active atoms to produce \( 1.55 \times 10^{14} \) radioactive decays every second.

Exam Tip: Be sure to write the units (atoms for count, Bq for activity) in your final answers to avoid losing easy points.

Question 895. The half life of a certain radioactive material against \alpha -decay is 100 days. After how much time, will the Undecayed fraction of the material be 6.25 % ? 
Answer:
We are given:
Half-life \( T = 100\text{ days} \)
Remaining fraction \( \frac{N}{N_0} = 6.25\% = \frac{6.25}{100} = \frac{1}{16} \)

The fraction of remaining radioactive nuclei after \( n \) half-lives is given by: \[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \] Substituting the given fraction: \[ \frac{1}{16} = \left(\frac{1}{2}\right)^n \] \[ \left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^n \]
\( \implies n = 4 \)
The total time \( t \) required is the number of half-lives multiplied by the half-life period: \[ t = n \times T = 4 \times 100\text{ days} = 400\text{ days} \]
Thus, it will take **400 days**.
In simple words: A fraction of 6.25% represents exactly \( 1/16 \) of the starting material. Since the sample halves every 100 days, it takes 4 half-lives (\( 1/2 \to 1/4 \to 1/8 \to 1/16 \)), which equals 400 days.

Exam Tip: Using the fraction formula \( \left(\frac{1}{2}\right)^n \) is the fastest and most reliable way to solve percentage-decay questions.

Question 896. The half life of radioactive substance is 20s. calculate- 
(i) The decay constant, and
(ii) time taken for the sample to decay 7/8th of the initial value.
Answer:
We are given:
Half-life \( T = 20\text{ s} \)

(i) **Decay Constant (\( \lambda \)):** \[ \lambda = \frac{0.693}{T} = \frac{0.693}{20\text{ s}} \approx 0.0346\text{ s}^{-1} \]
(ii) **Time Taken to decay \( 7/8 \)-th of the initial value:**
If \( 7/8 \)-th of the material has decayed, the remaining undecayed fraction is: \[ \frac{N}{N_0} = 1 - \frac{7}{8} = \frac{1}{8} \] Using the fraction relation: \[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \implies \frac{1}{8} = \left(\frac{1}{2}\right)^n \] \[ \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n \]
\( \implies n = 3 \)
The total time \( t \) is: \[ t = n \times T = 3 \times 20\text{ s} = 60\text{ s} \]
Thus, the decay constant is **\( 0.0346\text{ s}^{-1} \)** and the time taken is **\( 60\text{ s} \)**.
In simple words: (i) The decay constant is about \( 0.0346\text{ per second} \). (ii) To lose 7/8 of the material, only 1/8 must remain. This takes exactly 3 half-lives, which equals 60 seconds.

Exam Tip: Always read carefully whether the fraction given is the "decayed fraction" or the "remaining fraction" to avoid errors in your calculations.

Question 897. The activity of a radioactive element drops to \( \frac{1}{16} \)th of its initial value in 32 Years. Find the mean life of the sample. 
Answer:
We are given:
Remaining activity fraction \( \frac{R}{R_0} = \frac{1}{16} \)
Total time \( t = 32\text{ years} \)

The activity of a sample decays according to the same ratio as the number of nuclei: \[ \frac{R}{R_0} = \left(\frac{1}{2}\right)^n \] \[ \frac{1}{16} = \left(\frac{1}{2}\right)^n \implies \left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^n \]
\( \implies n = 4 \)
This means 4 half-lives have elapsed in 32 years. The half-life \( T \) is: \[ T = \frac{t}{n} = \frac{32\text{ years}}{4} = 8\text{ years} \]
The mean life \( \tau \) of the sample is: \[ \tau \approx 1.44 T = 1.44 \times 8\text{ years} \approx 11.52\text{ years} \]
Thus, the mean life of the sample is **\( 11.52\text{ years} \)**.
In simple words: The activity dropping to 1/16th means 4 half-lives have passed. Since this took 32 years, each half-life is 8 years. Multiplying by 1.44 gives a mean life of 11.52 years.

Exam Tip: First calculate the half-life from the activity ratio, then multiply by \( 1.44 \) to find the mean life.

Question 898. Calculate the energy release in MeV in the deuterium-tritium fusion reaction 
\( _1^2H + _1^3H \to _2^4He + _0^1n \)
Given \( m(_1^2H) = 2.014102\text{ u} \), \( m(_1^3H) = 3.016049\text{ u} \), \( m(_2^4He) = 4.002603\text{ u} \), \( m_n = 1.008665\text{ u} \) & \( 1\text{ u} = 931.5\text{ MeV}/c^2 \)
Answer:
The mass of the reactants before the fusion is: \[ m_i = m(_1^2H) + m(_1^3H) = 2.014102\text{ u} + 3.016049\text{ u} = 5.030151\text{ u} \]
The mass of the products after the fusion is: \[ m_f = m(_2^4He) + m_n = 4.002603\text{ u} + 1.008665\text{ u} = 5.011268\text{ u} \]
The mass defect \( \Delta m \) is: \[ \Delta m = m_i - m_f = 5.030151\text{ u} - 5.011268\text{ u} = 0.018883\text{ u} \]
The energy \( Q \) released in MeV is: \[ Q = \Delta m \times 931.5\text{ MeV} = 0.018883 \times 931.5\text{ MeV} \approx 17.59\text{ MeV} \]
Thus, the energy released in the reaction is **\( 17.59\text{ MeV} \)**.
In simple words: The products are slightly lighter than the starting isotopes by \( 0.018883\text{ u} \). This mass defect converts into \( 17.59\text{ MeV} \) of released energy.

Exam Tip: Be sure to perform your subtractions with all six decimal places to ensure your final MeV value is precise.

Question 899. Calculate the energy released if, \( ^{238}U \), emits an \( \alpha \)-particle. 
OR
Calculate the energy released in MeV in the following nuclear reaction. 
\( _{92}^{238}U \to _{90}^{234}Th + _2^4He + Q \)
[ Given, mass of \( _{92}^{238}U \) = 238.05079 u, mass of \( _{90}^{234}Th \) = 234.043630 u, mass of \( _2^4He \) = 4.002600 u & 1u = 931.5 MeV/\(c^2\) ]
Answer:
The mass of the parent nucleus Uranium-238 is: \[ m_i = 238.05079\text{ u} \]
The combined mass of the decay products (Thorium and the alpha particle) is: \[ m_f = m(Th) + m(He) = 234.043630\text{ u} + 4.002600\text{ u} = 238.046230\text{ u} \]
The mass defect \( \Delta m \) is: \[ \Delta m = m_i - m_f = 238.05079\text{ u} - 238.046230\text{ u} = 0.004560\text{ u} \]
The energy \( Q \) released in MeV is: \[ Q = \Delta m \times 931.5\text{ MeV} = 0.004560 \times 931.5\text{ MeV} \approx 4.25\text{ MeV} \]
Thus, the energy released in the reaction is **\( 4.25\text{ MeV} \)**.
In simple words: When Uranium-238 decays, it loses \( 0.00456\text{ u} \) of mass, which is converted and released as \( 4.25\text{ MeV} \) of kinetic energy carried by the products.

Exam Tip: Be sure to keep track of the number of decimal zeros during subtraction to prevent off-by-ten calculation errors.

Question 899a. A neutron is absorbed by a \( _3^6Li \) nucleus with the subsequent emission of an alpha particle. Write the corresponding nuclear reaction. Calculate the energy released in this nuclear reaction. 
OR
Calculate the energy released in the following nuclear reaction
\( _3^6Li + _0^1n \to _2^4He + _1^3H + Q \)
[ mass of \( _0^1n \) = 1.008665 u, mass of \( _3^6Li \) = 6.015126 u, mass of \( _2^4He \) = 4.002603 u, mass of \( _1^3H \) = 3.016049 u ]
Answer:
The written nuclear reaction is: \[ _3^6Li + _0^1n \to _2^4He + _1^3H + Q \]
The combined mass of the reactants is: \[ m_i = m(_3^6Li) + m_n = 6.015126\text{ u} + 1.008665\text{ u} = 7.023791\text{ u} \]
The combined mass of the products is: \[ m_f = m(_2^4He) + m(_1^3H) = 4.002603\text{ u} + 3.016049\text{ u} = 7.018652\text{ u} \]
The mass defect \( \Delta m \) is: \[ \Delta m = m_i - m_f = 7.023791\text{ u} - 7.018652\text{ u} = 0.005139\text{ u} \]
Using the conversion factor \( 1\text{ u} = 931\text{ MeV} \) (or standard \( 931.5\text{ MeV} \)): \[ Q = \Delta m \times 931\text{ MeV} = 0.005139 \times 931\text{ MeV} \approx 4.78\text{ MeV} \]
Thus, the energy released in the reaction is **\( 4.78\text{ MeV} \)**.
In simple words: The products Helium-4 and Tritium are lighter than the starting Lithium and neutron by \( 0.005139\text{ u} \), releasing \( 4.78\text{ MeV} \) of energy.

Exam Tip: Be sure to write out the requested nuclear equation first before performing the energy calculation.

Question 899b. (i) Write symbolically the nuclear \( \beta^+ \) decay process of \( _6^{11}C \). Is the decayed product X an isotope or isobar of \( _6^{11}C \) ?
(ii) Given tha mass value of \( _6^{11}C \) = 11.011434 and \( m(X) \) = 11.00935 . Estimate the Q value in this process. 
Answer:
(i) **Nuclear \( \beta^+ \) Decay Equation:**
The carbon decay process is: \[ _6^{11}C \to _5^{11}B + _{+1}^0e + \nu \] The decayed product is Boron-11 (\( _5^{11}B \)). Since both \( _6^{11}C \) and \( _5^{11}B \) share the same mass number (\( A = 11 \)) but have different atomic numbers, the product is an **isobar** of the parent carbon nucleus.

(ii) **Estimation of Q-value:**
*(Note: In \( \beta^+ \) decay, the Q-value calculation using atomic masses is \( Q = [M(C) - M(B) - 2m_e]c^2 \). If using the simplified mass difference given as \( \Delta m \)):* \[ \Delta m = m(_6^{11}C) - m(X) = 11.011434\text{ u} - 11.009350\text{ u} = 0.002084\text{ u} \]
Multiplying by the energy factor \( 931.5\text{ MeV} \): \[ Q = \Delta m \times 931.5\text{ MeV} = 0.002084 \times 931.5\text{ MeV} \approx 1.94\text{ MeV} \]
Thus, the Q-value is approximately **\( 1.94\text{ MeV} \)**.
In simple words: (i) Carbon-11 decays into Boron-11, which has the same mass weight, making it an isobar. (ii) The energy equivalent of the mass difference is \( 1.94\text{ MeV} \).

Exam Tip: Remember to specify "isobar" clearly, as definitions of isotopes, isobars, and isotones are frequently tested.

Question 899c. A nucleus \( _{10}^{23}Ne \), \( \beta^- \)-decays to give the nucleus of \( _{11}^{23}Na \). Write down the \( \beta^- \)-decay equation. Calculate the kinetic energy of electron emitted. (Rest mass of electron may be ignored.) 
(Given, \( m(_{10}^{23}Ne) = 22.994466\text{ u} \) & \( m(_{11}^{23}Na) = 22.989770\text{ u} \))
Answer:
**1. \( \beta^- \)-Decay Equation:** \[ _{10}^{23}Ne \to _{11}^{23}Na + _{-1}^0e + \bar{\nu} \]
**2. Calculation of Kinetic Energy (\( Q \)):**
Since the rest mass of the electron is ignored in this mass comparison, the mass defect \( \Delta m \) is: \[ \Delta m = m(_{10}^{23}Ne) - m(_{11}^{23}Na) = 22.994466\text{ u} - 22.989770\text{ u} = 0.004696\text{ u} \]
The energy \( Q \) released (which is carried as kinetic energy by the emitted electron and antineutrino) is: \[ Q = \Delta m \times 931.5\text{ MeV} = 0.004696 \times 931.5\text{ MeV} \approx 4.374\text{ MeV} \]
Thus, the maximum kinetic energy of the emitted electron is approximately **\( 4.374\text{ MeV} \)**.
In simple words: The decay of Neon-23 into Sodium-23 releases \( 4.374\text{ MeV} \) of energy, which is shared as kinetic energy between the electron and the antineutrino.

Exam Tip: Include the antineutrino (\( \bar{\nu} \)) in your decay equation to demonstrate a complete knowledge of beta-minus decay.

Question 899d. When a deuteron of mass 2.0141 u and negligible kinetic energy is absorbed by a Lithium \( _3^6Li \) nucleus of mass 6.0155 u, the compound nucleus disintegrates spontaneously in to two alpha particles each of mass 4.0026 u. Calculate the energy in Joules carried by each alpha particle. (\( 1\text{ u} = 1.66 \times 10^{-27}\text{ Kg} \))
Answer:
The reaction process is: \[ _3^6Li + _1^2H \to 2(_2^4He) + Q \]
The total mass of the reactants before the reaction is: \[ m_i = m(_3^6Li) + m(_1^2H) = 6.0155\text{ u} + 2.0141\text{ u} = 8.0296\text{ u} \]
The total mass of the two product alpha particles is: \[ m_f = 2 \times 4.0026\text{ u} = 8.0052\text{ u} \]
The mass defect \( \Delta m \) in atomic mass units is: \[ \Delta m = m_i - m_f = 8.0296\text{ u} - 8.0052\text{ u} = 0.0244\text{ u} \]
Converting this mass defect into kilograms: \[ \Delta m = 0.0244 \times 1.66 \times 10^{-27}\text{ kg} \approx 4.0504 \times 10^{-29}\text{ kg} \]
The total energy \( Q \) released in Joules is: \[ Q = \Delta m \cdot c^2 = (4.0504 \times 10^{-29}\text{ kg}) \times (3 \times 10^8\text{ m/s})^2 \] \[ Q = 4.0504 \times 10^{-29} \times 9 \times 10^{16}\text{ J} \approx 3.645 \times 10^{-12}\text{ J} \]
Since the total energy is shared equally between the two identical alpha particles, the energy carried by each alpha particle is: \[ E_{\alpha} = \frac{Q}{2} = \frac{3.645 \times 10^{-12}\text{ J}}{2} \approx 1.82 \times 10^{-12}\text{ J} \]
Thus, the energy carried by each alpha particle is **\( 1.82 \times 10^{-12}\text{ J} \)**.
In simple words: The reaction loses \( 0.0244\text{ u} \) of mass, releasing \( 3.645 \times 10^{-12}\text{ Joules} \) of energy. This energy is split equally, so each of the two alpha particles carries \( 1.82 \times 10^{-12}\text{ Joules} \).

Exam Tip: Be sure to divide the total energy by 2 at the end, as the question specifically asks for the energy carried by "each" of the two alpha particles.